Xi Coordinate

8/8/2019 Xi Coordinate

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Question 1:

Draw a quadrilateral in the Cartesian plane, whose vertices are (4, 5), (0, 7), (5, 5) and

(4, 2). Also, find its area.

Let ABCD be the given quadrilateral with vertices A (4, 5), B (0, 7), C (5, 5), and D (4, 2).

Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and

DA, the given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.

Accordingly, area (ABCD) = area (ABC) + area (ACD)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

Therefore, area of ABC


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Area of ACD

Thus, area (ABCD)

Question 2:

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid

point of the base is at the origin. Find vertices of the triangle.

Let ABC be the given equilateral triangle with side 2a.

Accordingly, AB = BC = CA = 2a

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B

are (0, a).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of

its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.


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On applying Pythagoras theorem to AOC, we obtain

(AC)2 = (OA)2 + (OC)2

(2a)2 = (OA)2 + a2

4a2 a2 = (OA)2

(OA)2 = 3a2

OA =

Coordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, a), (0, a), and or

(0, a), (0, a), and .

Question 3:

Find the distance between and when: (i) PQ is parallel to the y-axis,

(ii) PQ is parallel to the x-axis.

The given points are and .

(i) When PQ is parallel to the y-axis, x1 = x2.

In this case, distance between P and Q


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(ii) When PQ is parallel to the x-axis, y1 = y2.

In this case, distance between P and Q

Question 4:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain

a2 14a + 85 = a2 6a + 25

14a + 6a = 25 85

8a = 60

Thus, the required point on the x-axis is .

Question 5:

Find the slope of a line, which passes through the origin, and the mid-point of


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the line segment joining the points P (0, 4) and B (8, 0).

The coordinates of the mid-point of the line segment joining the points

P (0, 4) and B (8, 0) are

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and

(x2,y2) is given by .

Therefore, the slope of the line passing through (0, 0) and (4, 2) is

.

Hence, the required slope of the line is .

Question 6:

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (1, 1) are

the vertices of a right angled triangle.

The vertices of the given triangle are A (4, 4), B (3, 5), and C (1, 1).

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and

(x2,y2) is given by .

Slope of AB (m1)

Slope of BC (m2)

Slope of CA (m3)

It is observed that m1m3 = 1


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This shows that line segments AB and CA are perpendicular to each other

i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (1, 1) are the vertices of a right-angled triangle.

Question 7:

Find the slope of the line, which makes an angle of 30 with the positive direction ofy-

axis measured anticlockwise.

If a line makes an angle of 30 with the positive direction of the y-axis measured

anticlockwise, then the angle made by the line with the positive direction of the x-axis

measured anticlockwise is 90 + 30 = 120.

Thus, the slope of the given line is tan 120 = tan (1

Question 8:

Find the value ofx for which the points (x, 1), (2, 1) and (4, 5) are collinear.

If points A (x, 1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC


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Thus, the required value ofx is 1.

Question 9:

Without using distance formula, show that points (2, 1), (4, 0), (3, 3) and

(3, 2) are vertices of a parallelogram.

Let points (2, 1), (4, 0), (3, 3), and (3, 2) be respectively denoted by A, B, C, and D.

Slope of AB

Slope of CD =

Slope of AB = Slope of CD

AB and CD are parallel to each other.

Now, slope of BC =


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Slope of AD =

Slope of BC = Slope of AD

BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD

is a parallelogram.

Thus, points (2, 1), (4, 0), (3, 3), and (3, 2) are the vertices of a parallelogram.

Question 10:

Find the angle between the x-axis and the line joining the points (3, 1) and (4, 2).

The slope of the line joining the points (3, 1) and (4, 2) is

Now, the inclination () of the line joining the points (3, 1) and (4, 2) is given by

tan = 1

= (90 + 45) = 135

Thus, the angle between the x-axis and the line joining the points (3, 1) and (4, 2)

is 135.

Question 11:

The slope of a line is double of the slope of another line. If tangent of the angle between

them is , find the slopes of he lines.

Let be the slopes of the two given lines such that .

We know that ifisthe angle between the lines l1 and l2 with slopes m1 and m2, then

.


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It is given that the tangent of the angle between the two lines is .

Case I

Ifm = 1, then the slopes of the lines are 1 and 2.

Ifm = , then the slopes of the lines are and 1.

Case II


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Ifm = 1, then the slopes of the lines are 1 and 2.

Ifm = , then the slopes of the lines are .

Hence, the slopes of the lines are 1 and 2 or and 1 or1 and 2 or .

Question 12:

A line passes through .

If slope of the line is m, show that .

The slope of the line passing through is .

It is given that the slope of the line is m. Hence,

Question 13:If three point (h, 0), (a, b) and (0, k) lie on a line, show that .

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then



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On dividing both sides by kh, we obtain

Question 14:

Consider the given population and year graph. Find the slope of the line AB and using it,

find what will be the population in the year 2010?

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope

is

Let y be the population in the year 2010. Then, according to the given graph, line AB

must pass through point C (2010, y).



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Thus, the slope of line AB is , while in the year 2010, the population will be 104.5

crores.

Question 1:

Write the equations for the x and y-axes.

The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is y = 0.

Question 2:

Find the equation of the line which passes through the point (4, 3) with slope .

We know that the equation of the line passing through point , whose slope is m, is

.

Thus, the equation of the line passing through point (4, 3), whose slope is , is

Question 3:

Find the equation of the line which passes though (0, 0) with slope m.


.


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Thus, the equation of the line passing through point (0, 0), whose slope is m,is

(y 0) = m(x 0)

i.e.,y = mx

Question 4:

Find the equation of the line which passes though and is inclined with the x-axis

at an angle of 75.

The slope of the line that inclines with the x-axis at an angle of 75 is

m = tan 75


.

Thus, if a line passes though and inclines with the x-axis at an angle of 75, then

the equation of the line is given as

Question 5:


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Find the equation of the line which intersects the x-axis at a distance of 3 units to the left

of origin with slope 2.

It is known that if a line with slope m makes x-intercept d, then the equation of the line is

given as

y= m(x d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d= 3.

The slope of the line is given as m = 2

Thus, the required equation of the given line is

y = 2 [x (3)]

y = 2x 6

i.e., 2x + y + 6 = 0

Question 6:

Find the equation of the line which intersects the y-axis at a distance of 2 units above the

origin and makes an angle of 30 with the positive direction of the x-axis.

It is known that if a line with slope m makes y-intercept c, then the equation of the line is

given as

y= mx + c

Here, c = 2 and m = tan 30 .



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Question 7:

Find the equation of the line which passes through the points (1, 1) and (2, 4).

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

.

Therefore, the equation of the line passing through the points (1, 1) and

(2, 4) is

Question 8:

Find the equation of the line which is at a perpendicular distance of 5 units from the

origin and the angle made by the perpendicular with the positivex-axis is 30

Ifpis the length of the normal from the origin to a line and is the angle made by the

normal with the positive direction of the x-axis, then the equation of the line is given

by xcos + ysin = p.

Here, p = 5 units and = 30



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x cos 30 + y sin 30 = 5

Question 9:

The vertices of PQR are P (2, 1), Q (2, 3) and R (4, 5). Find equation of the median

through the vertex R.

It is given that the vertices of PQR are P (2, 1), Q (2, 3), and R (4, 5).

Let RL be the median through vertex R.

Accordingly, L is the mid-point of PQ.

By mid-point formula, the coordinates of point L are given by

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

.

Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and (x2,

y2) = (0, 2).

Hence,


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Thus, the required equation of the median through vertex R is .

Question 10:

Find the equation of the line passing through (3, 5) and perpendicular to the line through

the points (2, 5) and (3, 6).


We know that two non-vertical lines are perpendicular to each other if and only if their

slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (3, 6)

Now, the equation of the line passing through point (3, 5), whose slope is 5, is

Question 11:

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in

the ratio1

:n

. Find the equation of the line.

According to the section formula, the coordinates of the point that divides the line

segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by


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We know that two non-vertical lines are perpendicular to each other if and only if their

slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and

(2, 3)

Now, the equation of the line passing through and whose slope is is

given by

Question 12:

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes

through the point (2, 3).

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

It is given that the line cuts off equal intercepts on both the axes. This means that a = b.

Accordingly, equation (i) reduces to


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Since the given line passes through point (2, 3), equation (ii) reduces to

2 + 3 = a a = 5

On substituting the value ofa in equation (ii), we obtain

x + y = 5, which is the required equation of the line

Question 13:

Find equation of the line passing through the point (2, 2) and cutting off intercepts on the

axes whose sum is 9.

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

It is given thata + b = 9 b = 9 a (ii)

From equations (i) and (ii), we obtain

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to


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Ifa = 6 and b = 9 6 = 3, then the equation of the line is

Ifa = 3 and b = 9 3 = 6, then the equation of the line is

Question 14:

Find equation of the line through the point (0, 2) making an angle with the positivex-

axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2

units below the origin.

The slope of the line making an angle with the positive x-axis is

Now, the equation of the line passing through point (0, 2) and having a slope is

.


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The slope of line parallel to line is .

It is given that the line parallel to line crosses the y-axis 2 units below the

origin i.e., it passes through point (0, 2).

Hence, the equation of the line passing through point (0, 2) and having a slope is

Question 15:

The perpendicular from the origin to a line meets it at the point ( 2, 9), find the equation

of the line.

The slope of the line joining the origin (0, 0) and point (2, 9) is

Accordingly, the slope of the line perpendicular to the line joining the origin and point (2, 9) is

Now, the equation of the line passing through point (2, 9) and having a slope m2 is

Question 16:


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The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature

C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express

L in terms of C.

It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value

of L is 125.134.

Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between

L and C.

Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20,

124.942) and (110, 125.134) in the XY plane.

Therefore, the linear relation between L and C is the equation of the line passing through

points (20, 124.942) and (110, 125.134).

(L 124.942) =

Question 17:

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs

14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship

between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

The relationship between selling price and demand is linear.

Assuming selling price per litre along the x-axis and demand along the y-axis, we have

two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear

relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the

equation of the line passing through points (14, 980) and (16, 1220).


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When x = Rs 17/litre,

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Question 18:

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is

Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

Since P (a, b) is the mid-point of AB,


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Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).

The equation of the line passing through points (0, 2b) and (2a, 0) is

On dividing both sides by ab, we obtain

Thus,the equation of the line is .

Question 19:

Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of

the line.

Let AB be the line segment between the axes such that point R (h, k) divides AB in the

ratio 1: 2.

Let the respective coordinates of A and B be (x, 0) and (0, y).


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Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

Therefore, the respective coordinates of A and B are and (0, 3k).

Now, the equation of line AB passing through points and

(0, 3k) is

Thus,the required equation of the line is 2kx+ hy= 3hk.

Question 20:

By using the concept of equation of a line, prove that the three points (3, 0),

(2, 2) and (8, 2) are collinear.

In order to show that points (3, 0), (2, 2), and (8, 2) are collinear, it suffices to show

that the line passing through points (3, 0) and (2, 2) also passes through point (8, 2).

The equation of the line passing through points (3, 0) and (2, 2) is


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It is observed that at x = 8 and y = 2,

L.H.S. = 2 8 5 2 = 16 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (2, 2) also passes through point (8,

2). Hence, points (3, 0), (2, 2), and (8, 2) are collinear.

Question 1:

Reduce the following equations into slope-intercept form and find their slopes and the y-

intercepts.

(i) x+ 7y= 0 (ii) 6x+ 3y 5 = 0 (iii) y= 0

(i) The given equation is x+ 7y= 0.

It can be written as

This equation is of the form y = mx + c, where .

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept

are and 0 respectively.

(ii) The given equation is 6x + 3y 5 = 0.



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are2 and respectively.

(iii) The given equation is y = 0.


y = 0.x + 0 (3)

This equation is of the form y = mx + c, where m = 0 and c = 0.


are 0 and 0 respectively.

Question 2:

Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x+ 2y12 = 0 (ii) 4x 3y= 6 (iii) 3y+ 2 = 0.

(i) The given equation is 3x+ 2y12 = 0.


This equation is of the form , where a = 4 and b = 6.


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Therefore, equation (1) is in the intercept form, where the intercepts on the xand yaxes

are 4 and 6 respectively.

(ii) The given equation is 4x 3y= 6.


This equation is of the form , where a = and b = 2.

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes

are and 2 respectively.

(iii) The given equation is 3y+ 2 = 0.


This equation is of the form , where a = 0 and b = .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is

and it has no intercept on the x-axis.

Question 3:


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Reduce the following equations into normal form. Find their perpendicular distances

from the origin and angle between perpendicular and the positive x-axis.

(i) (ii) y 2 = 0 (iii) xy= 4

(i) The given equation is .

It can be reduced as:

On dividing both sides by , we obtain

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

x cos + y sin = p, we obtain = 120 andp = 4.

Thus, the perpendicular distance of the line from the origin is 4, while the angle between

the perpendicular and the positivex-axis is 120.

(ii) The given equation is y 2 = 0.

It can be reduced as 0.x + 1.y= 2

On dividing both sides by , we obtain 0.x + 1.y= 2

x cos 90 + ysin 90 = 2 (1)



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x cos + y sin = p, we obtain = 90 and p = 2.

Thus, the perpendicular distance of the line from the origin is 2, while the angle between

the perpendicular and the positivex-axis is 90.

(iii) The given equation is xy= 4.

It can be reduced as 1.x + (1) y= 4




x cos + y sin = p, we obtain = 315 and .

Thus, the perpendicular distance of the line from the origin is , while the angle

between the perpendicular and the positive x-axis is 315.

Question 4:

Find the distance of the point (1, 1) from the line 12(x+ 6) = 5(y 2).

The given equation of the line is 12(x+ 6) = 5(y 2).

12x + 72 = 5y 10

12x 5y + 82 = 0 (1)

On comparing equation (1) with general equation of line Ax+By+ C= 0, we obtain A =

12,B= 5, and C= 82.


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It is known that the perpendicular distance (d) of a line Ax+By+ C= 0 from a point

(x1, y1) is given by .

The given point is (x1, y1) = (1, 1).

Therefore, the distance of point (1, 1) from the given line

Question 5:

Find the points on the x-axis, whose distances from the line are 4 units.

The given equation of line is

On comparing equation (1) with general equation of line Ax+By+ C= 0, we obtain A =

4,B = 3, and C= 12.

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.



Therefore,


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Thus, the required points on the x-axis are (2, 0) and (8, 0).

Question 6:

Find the distance between parallel lines

(i) 15x+ 8y 34 = 0 and 15x+ 8y+ 31 = 0

(ii) l(x+ y) + p= 0 and l(x+ y) r= 0

It is known that the distance (d) between parallel lines Ax+By+ C1 = 0

and Ax+By+ C2 = 0 is given by .

(i) The given parallel lines are 15x+ 8y 34 = 0 and 15x+ 8y+ 31 = 0.

Here, A = 15,B = 8, C1 = 34, and C2 = 31.

Therefore, the distance between the parallel lines is

(ii) The given parallel lines are l(x+ y) + p= 0 and l(x+ y) r= 0.

lx+ ly + p= 0 and lx+ ly r= 0

Here, A = l,B = l, C1 = p, and C2 = r.


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Therefore, the distance between the parallel lines is

Question 7:

Find equation of the line parallel to the line 3x 4y + 2 = 0 and passing through the point

(2, 3).

The equation of the given line is

, which is of the form y = mx + c

Slope of the given line

It is known that parallel lines have the same slope.

Slope of the other line =

Now, the equation of the line that has a slope of and passes through the point (2, 3) is

Question 8:

Find equation of the line perpendicular to the line x 7y+ 5 = 0 and having xintercept 3.

The given equation of line is .


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The slope of the line perpendicular to the line having a slope of is

The equation of the line with slope 7 and x-intercept 3 is given by

y = m (x d)

y = 7 (x 3)

y = 7x + 21

7x + y = 21

Question 9:

Find angles between the lines

The given lines are .

The slope of line (1) is , while the slope of line (2) is .

The acute angle i.e., between the two lines is given by


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Thus, the angle between the given lines is either 30 or180 30 = 150.

Question 10:

The line through the points (h, 3) and (4, 1) intersects the line 7x 9y 19 = 0.at right

angle. Find the value ofh.

The slope of the line passing through points (h, 3) and (4, 1) is

The slope of line 7x 9y 19 = 0 or is .

It is given that the two lines are perpendicular.


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Thus, the value ofh is .

Question 11:

Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A

(x x1) +B (y y1) = 0.

The slope of line Ax + By + C = 0 or is

It is known that parallel lines have the same slope.

Slope of the other line =

The equation of the line passing through point (x1, y1) and having a slope is

Hence, the line through point (x1, y1) and parallel to line Ax + By + C = 0 is

A (x x1) + B (y y1) = 0

Question 12:


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Two lines passing through the point (2, 3) intersects each other at an angle of 60. If

slope of one line is 2, find equation of the other line.

It is given that the slope of the first line, m1 = 2.

Let the slope of the other line be m2.

The angle between the two lines is 60.

The equation of the line passing through point (2, 3) and having a slope of is


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In this case, the equation of the other line is .

The equation of the line passing through point (2, 3) and having a slope of is

In this case, the equation of the other line is .

Thus, the required equation of the other line is or

.

Question 13:

Find the equation of the right bisector of the line segment joining the points (3, 4) and (

1, 2).

The right bisector of a line segment bisects the line segment at 90.


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The end-points of the line segment are given as A (3, 4) and B (1, 2).

Accordingly, mid-point of AB

Slope of AB

Slope of the line perpendicular to AB =

The equation of the line passing through (1, 3) and having a slope of 2 is

(y 3) = 2 (x 1)

y 3 = 2x + 2

2x + y = 5

Thus, the required equation of the line is 2x +y = 5.

Question 14:

Find the coordinates of the foot of perpendicular from the point (1, 3) to the line 3x4y16 = 0.

Let (a, b) be the coordinates of the foot of the perpendicular from the point (1, 3) to the

line 3x 4y 16 = 0.

Slope of the line joining (1, 3) and (a, b), m1


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Slope of the line 3x 4y16 = 0 or

Since these two lines are perpendicular, m1m2 = 1

Point (a, b) lies on line 3x 4y = 16.

3a 4b = 16 (2)

On solving equations (1) and (2), we obtain

Thus, the required coordinates of the foot of the perpendicular are .

Question 15:

The perpendicular from the origin to the line y = mx + cmeets it at the point

(1, 2). Find the values ofm and c.

The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (1, 2).

Therefore, the line joining the points (0, 0) and (1, 2) is perpendicular to the given line.

Slope of the line joining (0, 0) and (1, 2)

The slope of the given line is m.


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Since point (

1, 2) lies on the given line, it satisfies the equation y = mx + c.

Thus, the respective values ofm and c are

Question 16:

Ifpand q are the lengths of perpendiculars from the origin to the lines xcos

ysin = kcos 2and xsec + ycosec = k, respectively, prove that p2 + 4q2 = k2

The equations of given lines are

x cos y sin= kcos 2 (1)

x sec+ y cosec = k (2)

The perpendicular distance (d) of a line Ax+By+ C= 0 from a point (x1, y1) is given by

.

On comparing equation (1) to the general equation of line i.e., Ax+By+ C= 0, we

obtain A = cos,B = sin, and C= kcos 2.

It is given that pis the length of the perpendicular from (0, 0) to line (1).

On comparing equation (2) to the general equation of line i.e., Ax+By+ C= 0, we

obtain A = sec,B = cosec, and C= k.


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It is given that q is the length of the perpendicular from (0, 0) to line (2).

From (3) and (4), we have

Hence, we proved that p2 + 4q2 = k2.

Question 17:

In the triangle ABC with vertices A (2, 3), B (4,1) and C (1, 2), find the equation and

length of altitude from the vertex A.

Let AD be the altitude of triangle ABC from vertex A.

Accordingly, ADBC


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The equation of the line passing through point (2, 3) and having a slope of1 is

(y 3) = 1(x 2)

x y + 1 = 0

yx = 1

Therefore, equation of the altitude from vertex A = yx = 1.

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is


.

On comparing equation (1) to the general equation of line Ax+By+ C= 0, we obtain A =

1,B=

1, and C= 3.

Length of AD

Thus, the equation and the length of the altitude from vertex A are y x = 1 and units

respectively.


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Question 18:

Ifpis the length of perpendicular from the origin to the line whose intercepts on the axes

are a andb, then show that .

It is known that the equation of a line whose intercepts on the axes are a and b is


.

On comparing equation (1) to the general equation of line Ax+By+ C= 0, we

obtain A = b,B= a, and C= ab.

Therefore, ifpis the length of the perpendicular from point (x1, y1) = (0, 0) to line (1), we

obtain

On squaring both sides, we obtain


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Hence, we showed that .

Question 1:

Find the values ofkfor which the line is

(a) Parallel to the x-axis,

(b) Parallel to the y-axis,

(c) Passing through the origin.

The given equation of line is

(k 3) x (4 k2) y + k2 7k+ 6 = 0 (1)

(a) If the given line is parallel to the x-axis, then

Slope of the given line = Slope of the x-axis

The given line can be written as

(4 k2) y = (k 3) x + k2 7k+ 6 = 0

, which is of the form y = mx + c.

Slope of the given line =

Slope of the x-axis = 0

Thus, if the given line is parallel to the x-axis, then the value ofkis 3.


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(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be

undefined.

The slope of the given line is .

Now, is undefined atk2 = 4

k2 = 4

k= 2

Thus, if the given line is parallel to the y-axis, then the value ofkis 2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the

given equation of line.

Thus, if the given line is passing through the origin, then the value ofkis either1 or 6.

Question 2:

Find the values ofand p, if the equation is the normal form of the line

.

The equation of the given line is .

This equation can be reduced as


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On comparing equation (1) to , we obtain

Since the values of sin and cos are negative,

Thus, the respective values ofand p are and 1

Question 3:

Find the equations of the lines, which cut-off intercepts on the axes whose sum and

product are 1 and 6, respectively.

Let the intercepts cut by the given lines on the axes be a and b.

It is given that

a + b = 1 (1)

ab = 6 (2)


a = 3 and b = 2 ora = 2 and b = 3

It is known that the equation of the line whose intercepts on the axes are a and b is


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Case I:a = 3 and b = 2

In this case, the equation of the line is 2x + 3y + 6 = 0, i.e., 2x 3y = 6.

Case II:a = 2 and b = 3

In this case, the equation of the line is 3x 2y + 6 = 0, i.e., 3x + 2y = 6.

Thus, the required equation of the lines are 2x 3y = 6 and 3x + 2y = 6.

Question 4:

What are the points on the y-axis whose distance from the line is 4 units.

Let (0, b) be the point on the y-axis whose distance from line is 4 units.

The given line can be written as 4x + 3y 12 = 0 (1)

On comparing equation (1) to the general equation of line Ax+By+ C= 0, we obtain A =

4,B= 3, and C= 12.



Therefore, if (0, b) is the point on the y-axis whose distance from line is 4 units,

then:


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Thus, the required points are and .

Question 5:

Find the perpendicular distance from the origin to the line joining the

points

The equation of the line joining the points is given by



Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is


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Question 6:

Find the equation of the line parallel to y-axis and drawn through the point of intersection

of the lines x 7y + 5 = 0 and 3x + y = 0.

The equation of any line parallel to the y-axis is of the form

x = a (1)

The two given lines are

x 7y + 5 = 0 (2)

3x + y = 0 (3)

On solving equations (2) and (3), we obtain .


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Therefore, is the point of intersection of lines (2) and (3).

Since line x = a passes through point , .

Thus, the required equation of the line is .

Question 7:

Find the equation of a line drawn perpendicular to the line through the point,

where it meets the y-axis.

The equation of the given line is .

This equation can also be written as 3x + 2y 12 = 0



Slope of line perpendicular to the given line

Let the given line intersect the y-axis at (0, y).

On substituting x with 0 in the equation of the given line, we obtain

The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of and passes through point (0, 6) is


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Thus, the required equation of the line is

Question 8:

Find the area of the triangle formed by the lines y x= 0, x + y = 0 and x k= 0.

The equations of the given lines are

y x = 0 (1)

x + y = 0 (2)

x k= 0 (3)

The point of intersection of lines (1) and (2) is given by

x = 0 andy = 0


x = kand y = k


x = kandy= k

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, k), and

(k, k).

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

.

Therefore, area of the triangle formed by the three given lines


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Question 9:

Find the value ofp so that the three lines 3x +y 2 = 0, px + 2y 3 = 0 and 2x y 3 = 0

may intersect at one point.


3x + y 2 = 0 (1)

px+ 2y 3 = 0 (2)

2xy 3 = 0 (3)


x= 1 and y = 1

Since these three lines may intersect at one point, the point of intersection of lines (1) and

(3) will also satisfy line (2).

p (1) + 2 (1) 3 = 0

p 2 3 = 0

p = 5

Thus, the required value ofp is 5.

Question 10:

If three lines whose equations are

concurrent, then show that


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y = m1x + c1 (1)

y = m2x + c2 (2)

y = m3x + c3 (3)

On subtracting equation (1) from (2), we obtain

On substituting this value ofx in (1), we obtain

is the point of intersection of lines (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of

lines (1) and (2) will also satisfy equation (3).


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Hence,

Question 11:

Find the equation of the lines through the point (3, 2) which make an angle of 45 with

the line x 2y = 3.

Let the slope of the required line be m1.

The given line can be written as , which is of the form y = mx + c

Slope of the given line =

It is given that the angle between the required line and line x 2y = 3 is 45.

We know that ifisthe acute angle between lines l1 and l2 with

slopes m1 and m2 respectively, then .


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Case I:m1 = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y 2 = 3 (x 3)

y 2 = 3x 9

3x y = 7

Case II:m1 =

The equation of the line passing through (3, 2) and having a slope of is:


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Thus, the equations of the lines are 3x y = 7 and x + 3y = 9.

Question 12:

Find the equation of the line passing through the point of intersection of the lines 4x +

7y 3 = 0 and 2x 3y + 1 = 0 that has equal intercepts on the axes.

Let the equation of the line having equal intercepts on the axes be

On solving equations 4x + 7y 3 = 0 and 2x 3y + 1 = 0, we obtain .

is the point of intersection of the two given lines.

Since equation (1) passes through point ,

Equation (1

) becomes

Thus, the required equation of the line is

Question 14:

In what ratio, the line joining (1, 1) and (5, 7) is divided by the line


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x + y = 4?

The equation of the line joining the points (1, 1) and (5, 7) is given by


x + y 4 = 0 (2)


x = 1 and y = 3

Let point (1, 3) divide the line segment joining (1, 1) and (5, 7) in the ratio 1:k.

Accordingly, by section formula,

Thus, the line joining the points (1, 1) and (5, 7) is divided by line

x + y = 4 in the ratio 1:2.

Question 13:


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Show that the equation of the line passing through the origin and making an angle with

the line .

Let the equation of the line passing through the origin be y = m1x.

If this line makes an angle ofwith line y = mx + c, then angle is given by

Case I:


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Case II:

Therefore, the required line is given by

Question 15:

Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x y = 0.

The given lines are

2x y = 0 (1)

4x + 7y + 5 = 0 (2)

A (1, 2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).


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Coordinates of point B are .

By using distance formula, the distance between points A and B can be obtained as

Thus, the required distance is .

Question 16:

Find the direction in which a straight line must be drawn through the point (1, 2) so thatits point of intersection with the line x + y = 4 may be at a distance of 3 units from this

point.

Let y = mx + c be the line through point (1, 2).

Accordingly, 2 = m (1) + c.


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2 = m + c

c= m + 2

y= mx + m + 2 (1)

The given line is

x + y = 4 (2)


is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (1, 2), according to distance

formula,

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-

axis.

Question 18:


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Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to

be a plane mirror.


x + 3y = 7 (1)

Let point B (a, b) be the image of point A (3, 8).

Accordingly, line (1) is the perpendicular bisector of AB.

Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have


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On solving equations (2) and (3), we obtain a = 1 and b = 4.

Thus, the image of the given point with respect to the given line is (1, 4).

uestion 19:

If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the

value ofm.


y = 3x + 1 (1)

2y = x+ 3 (2)

y = mx + 4 (3)

Slope of line (1), m1 = 3

Slope of line (2),

Slope of line (3), m3 = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).


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Thus, the required value ofm is

Question 20:

If sum of the perpendicular distances of a variable point P (x,y) from the lines x + y 5 =

0 and 3x 2y + 7 = 0 is always 10. Show that P must move on a line.


x + y 5 = 0 (1)

3x 2y + 7 = 0 (2)

The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by

It is given that .

, which is the equation of a

line.

Similarly, we can obtain the equation of line for any signs of .

Thus, point P must move on a line.


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Question 21:

Find equation of the line which is equidistant from parallel lines 9x+ 6y 7 = 0 and 3x +

2y + 6 = 0.


9x + 6y 7 = 0 (1)

3x + 2y + 6 = 0 (2)

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The

perpendicular distance of P (h, k) from line (1) is given by

The perpendicular distance of P (h,k) from line (2) is given by

Since P (h, k) is equidistant from lines (1) and (2),

9h + 6k 7 = 9h 6k18

18h + 12k+ 11 = 0


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Thus, the required equation of the line is 18x + 12y + 11 = 0.

Question 22:

A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the

reflected ray passes through the point (5, 3). Find the coordinates of A.

Let the coordinates of point A be (a, 0).

Draw a line (AL) perpendicular to the x-axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

BAL = CAL =

Let CAX =

OAB = 180 (+ 2) = 180 [+ 2(90 )]

= 180 180 + 2

=

BAX = 180


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From equations (1) and (2), we obtain

Thus, the coordinates of point A are .

Question 23:

Prove that the product of the lengths of the perpendiculars drawn from the points


Length of the perpendicular from point to line (1) is


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Length of the perpendicular from point to line (2) is

On multiplying equations (2) and (3), we obtain

Hence, proved.

Question 24:


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A person standing at the junction (crossing) of two straight paths represented by the

equations 2x 3y + 4 = 0 and 3x + 4y 5 = 0 wants to reach the path whose equation is

6x 7y + 8 = 0 in the least time. Find equation of the path that he should follow.


2x 3y + 4 = 0 (1)

3x + 4y 5 = 0 (2)

6x 7y + 8 = 0 (3)

The person is standing at the junction of the paths represented by lines (1) and (2).


Thus, the person is standing at point .

The person can reach path (3) in the least time if he walks along the perpendicular line to

(3) from point .

Slope of the line perpendicular to line (3)

The equation of the line passing through and having a slope of is given

by

Hence, the path that the person should follow is


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Circles

Question 1:

Find the equation of the circle with centre (0, 2) and radius 2

The equation of a circle with centre (h, k) and radius ris given as

(xh)2 + (y k)2 = r2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x 0)2 + (y 2)2 = 22

x2 + y2 + 4 4 y = 4

x2 + y2 4y = 0

Question 2:

Find the equation of the circle with centre (2, 3) and radius 4


(xh)2 + (y k)2 = r2

It is given that centre (h, k) = (2, 3) and radius (r) = 4.


(x + 2)2 + (y 3)2 = (4)2

x2 + 4x + 4 + y2 6y + 9 = 16

x2 + y2 + 4x 6y 3 = 0

Question 3:

Find the equation of the circle with centre and radius


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(xh)2 + (y k)2 = r2

It is given that centre (h, k) = and radius (r) = .


Question 4:

Find the equation of the circle with centre (1, 1) and radius


(xh)2 + (y k)2 = r2

It is given that centre (h, k) = (1, 1) and radius (r) = .


Question 5:


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Find the equation of the circle with centre (a, b) and radius


(xh)2

+ (y k)2

= r2

It is given that centre (h, k) = (a, b) and radius (r) = .


Question 6:

Find the centre and radius of the circle (x + 5)2 + (y 3)2 = 36

The equation of the given circle is (x + 5)2 + (y 3)2 = 36.

(x + 5)2 + (y 3)2 = 36

{x (5)}2 + (y 3)2 = 62, which is of the form (x h)2 + (y k)2 = r2, where h = 5, k=

3, and r= 6.

Thus, the centre of the given circle is (5, 3), while its radius is 6.

Question 7:

Find the centre and radius of the circle x2 + y2 4x 8y 45 = 0

The equation of the given circle is x2 + y2 4x 8y 45 = 0.

x2 + y2 4x 8y 45 = 0

(x2 4x) + (y2 8y) = 45

{x2 2(x)(2) + 22} + {y2 2(y)(4)+ 42} 4 16 = 45

(x 2)2 + (y 4)2 = 65


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(x 2)2 + (y 4)2 = , which is of the form (x h)2 + (y k)2 = r2, where h =

2, k= 4, and .

Thus, the centre of the given circle is (2, 4), while its radius is .

Question 8:

Find the centre and radius of the circle x2 + y2 8x + 10y 12 = 0

The equation of the given circle is x2 + y2 8x + 10y 12 = 0.

x2 + y2 8x + 10y 12 = 0

(x2 8x) + (y2+ 10y) = 12

{x2 2(x)(4) + 42} + {y2+ 2(y)(5) + 52}16 25 = 12

(x 4)2 + (y + 5)2 = 53

,which is of the form (x h)2 + (y k)2 = r2, where h =

4, k= 5, and .

Thus, the centre of the given circle is (4, 5), while its radius is .

Question 9:

Find the centre and radius of the circle 2x2 + 2y2 x = 0

The equation of the given circle is 2x2 + 2y2 x = 0.


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, which is of the form (x h)2 + (y k)2 = r2, where h = , k=

0, and .

Thus, the centre of the given circle is , while its radius is .

Question 10:

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose

centre is on the line 4x + y = 16.

Let the equation of the required circle be (x h)2 + (y k)2 = r2.

Since the circle passes through points (4, 1) and (6, 5),

(4 h)2 + (1 k)2 = r2 (1)

(6 h)2 + (5 k)2 = r2 (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k= 16 (3)


(4 h)2 + (1 k)2 = (6 h)2 + (5 k)2

16 8h + h2 + 1 2k+ k2 = 36 12h + h2 + 25 10k+ k2

16 8h + 1 2k= 36 12h + 25 10k

4h + 8k= 44

h + 2k= 11 (4)

On solving equations (3) and (4), we obtain h = 3 and k= 4.

On substituting the values ofh and kin equation (1), we obtain


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(4 3)2 + (1 4)2 = r2

(1)2 + ( 3)2 = r2

1 + 9 = r2

r2 = 10

Thus, the equation of the required circle is

(x 3)2 + (y 4)2 =

x2 6x + 9 + y2 8y + 16 = 10

x2 + y2 6x 8y + 15 = 0

Question 11:

Find the equation of the circle passing through the points (2, 3) and (1, 1) and whose

centre is on the line x 3y 11 = 0.


Since the circle passes through points (2, 3) and (1, 1),

(2 h)2 + (3 k)2 = r2 (1)

(1 h)2 + (1 k)2 = r2 (2)

Since the centre (h, k) of the circle lies on line x 3y 11 = 0,

h 3k= 11 (3)


(2 h)2+ (3 k)2 = (1 h)2 + (1 k)2

4 4h + h2 + 9 6k+ k2 = 1 + 2h + h2 + 1 2k+ k2


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4 4h + 9 6k= 1 + 2h + 1 2k

6h + 4k= 11 (4)


On substituting the values ofh and kin equation (1), we obtain


Question 12:

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes

through the point (2, 3).


Since the radius of the circle is 5 and its centre lies on the x-axis, k= 0 and r= 5.


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Now, the equation of the circle becomes (x h)2 + y2 = 25.

It is given that the circle passes through point (2, 3).

When h = 2, the equation of the circle becomes

(x + 2)2 + y2 = 25

x2 + 4x+ 4 + y2 = 25

x2 + y2 + 4x 21 = 0

When h = 6, the equation of the circle becomes

(x 6)2 + y2 = 25

x2 12x +36 + y2 = 25

x2 + y2 12x + 11 = 0

Question 13:

Find the equation of the circle passing through (0, 0) and making intercepts a and b on

the coordinate axes.


Since the centre of the circle passes through (0, 0),

(0 h)2 + (0 k)2 = r2

h2 + k2 = r2


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The equation of the circle now becomes (x h)2 + (y k)2 = h2 + k2.

It is given that the circle makes intercepts a and b on the coordinate axes. This means that

the circle passes through points (a, 0) and (0, b). Therefore,

(a h)2 + (0 k)2 = h2 + k2 (1)

(0 h)2 + (b k)2 = h2 + k2 (2)

From equation (1), we obtain

a2 2ah + h2 + k2 = h2 + k2

a2 2ah = 0

a(a 2h) = 0

a = 0 or (a 2h) = 0

However, a 0; hence, (a 2h) = 0 h = .

From equation (2), we obtain

h2 + b2 2bk+ k2 = h2 + k2

b2 2bk= 0

b(b 2k) = 0

b = 0 or(b 2k) = 0

However, b 0; hence, (b 2k) = 0k= .



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Question 14:

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance

between the points (2, 2) and (4, 5).

Thus, the equation of the circle is

Question 15:

Does the point (2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

The equation of the given circle is x2 + y2 = 25.

x2 + y2 = 25

(x 0)2 + (y 0)2 = 52, which is of the form (x h)2 + (y k)2 = r2, where h = 0, k= 0,

andr= 5.

Centre = (0, 0) and radius = 5


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Distance between point (2.5, 3.5) and centre (0, 0)

Since the distance between point (2.5, 3.5) and centre (0, 0) of the circle is less than the

radius of the circle, point (2.5, 3.5) lies inside the circle.


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Conic Sections

Question 1:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the

length of the latus rectum fory2

=12x

The given equation is y2 = 12x.

Here, the coefficient ofxis positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain

4a = 12 a = 3

Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of direcctrix, x = a i.e., x= 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 3 = 12

Question 2:


length of the latus rectum forx2 = 6y

The given equation is x2 = 6y.

Here, the coefficient ofyis positive. Hence, the parabola opens upwards.

On comparing this equation with x2 = 4ay, we obtain

Coordinates of the focus = (0, a) =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,


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Length of latus rectum = 4a = 6

Question 3:


length of the latus rectum fory2 = 8x


Here, the coefficient ofxis negative. Hence, the parabola opens towards the left.


4a = 8 a = 2

Coordinates of the focus = (a, 0) = (2, 0)


Equation of directrix, x = a i.e.,x = 2


Question 4:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and thelength of the latus rectum forx2 = 16y


Here, the coefficient ofyis negative. Hence, the parabola opens downwards.


4a = 16 a = 4

Coordinates of the focus = (0, a) = (0, 4)


Equation of directrix, y = a i.e., y = 4


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Question 5:


length of the latus rectum fory2 = 10x


Here, the coefficient ofxis positive. Hence, the parabola opens towards the right.


Coordinates of the focus = (a, 0)




Question 6:


length of the latus rectum forx2 = 9y


Here, the coefficient ofyis negative. Hence, the parabola opens downwards.


Coordinates of the focus =



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Question 7:

Find the equation of the parabola that satisfies the following conditions: Focus (6, 0);

directrix x = 6

Focus (6, 0); directrix, x = 6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y2 = 4ax or

y2 = 4ax.

It is also seen that the directrix, x = 6 is to the left of the y-axis, while the focus (6, 0) is

to the right of the y-axis. Hence, the parabola is of the form y2 = 4ax.

Here, a = 6

Thus, the equation of the parabola is y2 = 24x.

Question 8:

Find the equation of the parabola that satisfies the following conditions: Focus (0, 3);

directrix y= 3

Focus = (0, 3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x2 = 4ay or

x2= 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus

(0, 3) is below the x-axis. Hence, the parabola is of the form x2 = 4ay.


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Here, a = 3

Thus, the equation of the parabola is x2 = 12y.

Question 9:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0);

focus (3, 0)

Vertex (0, 0); focus (3, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is

the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y2 = 4 3 x, i.e., y2 = 12x

Question 10:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0)

focus (2, 0)

Vertex (0, 0) focus (2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axisis the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (2, 0), a = 2.

Thus, the equation of the parabola is y2 = 4(2)x, i.e., y2 = 8x

Question 11:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0)

passing through (2, 3) and axis is along x-axis

Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the

parabola is either of the form y2 = 4ax ory2 = 4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.


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Therefore, the equation of the parabola is of the form y2 = 4ax, while point

(2, 3) must satisfy the equation y2 = 4ax.

Thus, the equation of the parabola is

Question 12:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0),

passing through (5, 2) and symmetric with respect to y-axis

Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of

the parabola is either of the form x2 = 4ay orx2 = 4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form x2 = 4ay, while point

(5, 2) must satisfy the equation x2 = 4ay.

Thus, the equation of the parabola is

Question 1:


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Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the

eccentricity and the length of the latus rectum of the ellipse

The given equation is .

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain a = 6 and b = 4.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (6, 0) and (6, 0).

Length of major axis = 2a = 12

Length of minor axis = 2b = 8

Length of latus rectum

Question 2:




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Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with , we obtain b = 2 and a = 5.

Therefore,


The coordinates of the vertices are (0, 5) and (0, 5)




Question 3:





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Therefore,


The coordinates of the vertices are .




Question 4:







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Therefore,


The coordinates of the vertices are (0, 10).




Question 5:








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Therefore,


The coordinates of the vertices are ( 7, 0).




Question 6:







Therefore,



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The coordinates of the vertices are (0, 20)




Question 7:


eccentricity and the length of the latus rectum of the ellipse 36x2 + 4y2 = 144

The given equation is 36x2 + 4y2 = 144.




On comparing equation (1) with , we obtain b = 2 and a = 6.

Therefore,



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Question 8:


eccentricity and the length of the latus rectum of the ellipse 16x2 + y2 = 16

The given equation is 16x2 + y2 = 16.




On comparing equation (1) with , we obtain b = 1 and a = 4.

Therefore,



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Question 9:


eccentricity and the length of the latus rectum of the ellipse 4x2 + 9y2 = 36

The given equation is 4x2 + 9y2 = 36.





Therefore,



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The coordinates of the vertices are ( 3, 0).




Question 10:

Find the equation for the ellipse that satisfies the given conditions: Vertices ( 5, 0), foci

( 4, 0)

Vertices ( 5, 0), foci ( 4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-

major axis.

Accordingly, a = 5 and c = 4.

It is known that .

Thus, the equation of the ellipse is

Question 11:

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, 13), foci

(0, 5)

Vertices (0, 13), foci (0, 5)


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Here, the vertices are on the y-axis.


major axis.

Accordingly, a = 13 and c = 5.

It is known that .


Question 12:

Find the equation for the ellipse that satisfies the given conditions: Vertices ( 6, 0), foci

( 4, 0)

Vertices ( 6, 0), foci ( 4, 0)

Here, the vertices are on the x-axis.


major axis.

Accordingly, a = 6, c = 4.

It is known that .


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Question 13:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis

( 3, 0), ends of minor axis (0, 2)

Ends of major axis ( 3, 0), ends of minor axis (0, 2)

Here, the major axis is along the x-axis.


major axis.

Accordingly, a = 3 and b = 2.

Thus, the equation of the ellipse is .

Question 14:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis

, ends of minor axis ( 1, 0)

Ends of major axis , ends of minor axis ( 1, 0)

Here, the major axis is along the y-axis.


major axis.

Accordingly, a = and b = 1.



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Question 15:

Find the equation for the ellipse that satisfies the given conditions: Length of major axis

26, foci ( 5, 0)

Length of major axis = 26; foci = ( 5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.


major axis.

Accordingly, 2a = 26 a = 13 and c = 5.

It is known that .


Question 16:

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis

16, foci (0, 6)

Length of minor axis = 16; foci = (0, 6).

Since the foci are on the y-axis, the major axis is along the y-axis.


major axis.

Accordingly, 2b = 16 b = 8 and c = 6. It is known that .



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Question 17:

Find the equation for the ellipse that satisfies the given conditions: Foci ( 3, 0), a = 4

Foci ( 3, 0), a = 4



major axis.

Accordingly, c = 3 and a = 4. It is known that .


Question 18:

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at

the origin; foci on the xaxis.

It is given that b = 3, c = 4, centre at the origin; foci on the xaxis.



major axis.

Accordingly, b = 3, c = 4.

It is known that .


Question 19:


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Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major

axis on the y-axis and passes through the points (3, 2) and (1, 6).

Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse

will be of the form

The ellipse passes through points (3, 2) and (1, 6). Hence,

On solving equations (2) and (3), we obtain b2 = 10 and a2 = 40.


Question 20:

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Since the major axis is on the x-axis, the equation of the ellipse will be of the form

The ellipse passes through points (4, 3) and (6, 2). Hence,

On solving equations (2) and (3), we obtain a2 = 52 and b2 = 13.


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Question 1:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the

latus rectum of the hyperbola


On comparing this equation with the standard equation of hyperbola i.e., , we

obtain a= 4 and b = 3.

We know that a2 + b2 = c2.

Therefore, The coordinates of the foci are ( 5, 0). The coordinates of the vertices are

( 4, 0).


Question 2:


latus rectum of the hyperbola

The given equation is . On comparing this equation with

the standard equation of hyperbola i.e., , we obtain a= 3 and .



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Therefore, The coordinates of the foci are ( 10, 0). The coordinates of the vertices are

( 6, 0).


Question 5:


latus rectum of the hyperbola 5y2 9x2 = 36

The given equation is 5y2 9x2 = 36.

On comparing equation (1) with the standard equation of hyperbola i.e., , we

obtain a= and b = 2. We know that a2 + b2 = c2.

Therefore, the coordinates of the foci are . The coordinates of the vertices are

.


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Question 6:


latus rectum of the hyperbola 49y2 16x2 = 784

The given equation is 49y2 16x2 = 784. It can be written as 49y2 16x2 = 784

On comparing equation (1) with the standard equation

of hyperbola i.e., , we obtain a= 4 and b = 7.

We know that a2 + b2 = c2. Therefore, The coordinates of the foci

are . The coordinates of the vertices are (0, 4).


Question 7:

Find the equation of the hyperbola satisfying the give conditions: Vertices ( 2, 0), foci

( 3, 0) Vertices ( 2, 0), foci ( 3, 0) Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are ( 2, 0), a = 2. Since the foci are ( 3, 0), c = 3.


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We know that a2 + b2 = c2. Thus, the equation of the hyperbola is

.

Question 8:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, 5), foci

(0, 8)

Vertices (0, 5), foci (0, 8) Here, the vertices are on the y-axis.


Since the vertices are (0, 5), a = 5. Since the foci are (0, 8), c = 8.


Thus, the equation of the hyperbola is

Question 9:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, 3), foci

(0, 5) Vertices (0, 3), foci (0, 5) Here, the vertices are on the y-axis.


Since the vertices are (0, 3), a = 3. Since the foci are (0, 5), c = 5.

We know that a2 + b2 = c2. 32 + b2 = 52 b2 = 25 9 = 16


Question 10:


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Find the equation of the hyperbola satisfying the give conditions: Foci ( 5, 0), the

transverse axis is of length 8.

Foci ( 5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis. Therefore, the equation of the hyperbola is of the form

. Since the foci are ( 5, 0), c = 5. Since the length of the transverse axis is

8, 2a = 8a = 4.

We know that a2 + b2 = c2. 42 + b2 = 52 b2 = 25 16 = 9


Question 11:

Find the equation of the hyperbola satisfying the give conditions: Foci (0, 13), the

conjugate axis is of length 24.

Foci (0, 13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.


Since the foci are (0, 13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 b = 12.

We know that a2 + b2 = c2. a2 + 122 = 132 a2 = 169 144 = 25



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Question 12:

Find the equation of the hyperbola satisfying the give conditions: Foci , the

latus rectum is of length 8.

Foci , the latus rectum is of length 8.

Here, the foci are on the x-axis. Therefore, the equation of the hyperbola is of the form

.

Since the foci are , c = . Length of latus rectum = 8

We know that a2 + b2 = c2. a2 + 4a = 45 a2 + 4a 45 = 0

a2 + 9a 5a 45 = 0 (a + 9) (a 5) = 0

a = 9, 5 Since a is non-negative, a = 5.

b2 = 4a = 4 5 = 20 Thus, the equation of the hyperbola is

Question 13:

Find the equation of the hyperbola satisfying the give conditions: Foci ( 4, 0), the latus

rectum is of length 12

Foci ( 4, 0), the latus rectum is of length 12. Here, the foci are on the x-axis.


Since the foci are ( 4, 0), c = 4. Length of latus rectum = 12


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We know that a2 + b2 = c2. a2 + 6a = 16 a2 + 6a 16 = 0

a2 + 8a 2a 16 = 0 (a + 8) (a 2) = 0 a = 8, 2

Since a is non-negative, a = 2.

b2 = 6a = 6 2 = 12 Thus, the equation of the hyperbola is

Question 14:Find the equation of the hyperbola satisfying the give conditions: Vertices

( 7, 0),

Vertices ( 7, 0), Here, the vertices are on thex-axis.


Since the vertices are ( 7, 0), a = 7. It is given that




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Question 15:

Find the equation of the hyperbola satisfying the give conditions: Foci ,

passing through (2, 3)

Foci , passing through (2, 3)

Here, the foci are on the y-axis.Therefore, the equation of the hyperbola is of the form

.

Since the foci are , c = .

We know that a2 + b2 = c2. a2 + b2 = 10 b2 = 10 a2 (1)

Since the hyperbola passes through point (2, 3),


In hyperbola, c > a, i.e., c2 > a2

a2 = 5

b2 = 10 a2 = 10 5 = 5


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Question 2: An arch is in the form of a parabola with its axis vertical. The arch is 10 m

high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its

vertical axis is along the positive y-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

It can be clearly seen that the parabola passes through point .

Therefore, the arch is in the form of a parabola whose equation is .

When y = 2 m,


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Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately

2.23 m.

Question 3:

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The

roadway which is horizontal and1

00 m long is supported by vertical wires attached to thecable, the longest wire being 30 m and the shortest being 6 m. Find the length of a

supporting wire attached to the roadway 18 m from the middle.

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken

as the vertex of the parabola, while its vertical axis is taken along the positive y-axis. This

can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the

cable.

DF is the supporting wire attached to the roadway, 18 m from the middle.

Here, AB = 30 m, OC = 6 m, and .

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).


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The coordinates of point A are (50, 30 6) = (50, 24).

Since A (50, 24) is a point on the parabola,

Equation of the parabola, or 6x2 = 625y

Thex-coordinate of point D is 18.

Hence, at x = 18,

DE = 3.11 m

DF = DE + EF = 3.11 m + 6 m = 9.11 m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is

approximately 9.11 m.

Question 4:

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find

the height of the arch at a point 1.5 m from one end.

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is

clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major

axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically

represented as


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The equation of the semi-ellipse will be of the form , where a is the

semi-major axis

Accordingly, 2a = 8 a = 4

b = 2

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC OB.

OA = (4 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value ofx with 2.5 in equation (1), we obtain


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AC = 1.56 m

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Question 5:

A rod of length 12 cm moves with its ends always touching the coordinate axes.

Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in

contact with the x-axis.

Let AB be the rod making an angle with OX and P (x,y) be the point on it such that AP

= 3 cm.

Then, PB = AB AP = (12 3) cm = 9 cm [AB = 12 cm]

From P, draw PQOY and PROX.

In PBQ,


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In PRA,

Thus, the equation of the locus of point P on the rod is

Question 6:

Find the area of the triangle formed by the lines joining the vertex of the parabola x2 =

12y to the ends of its latus rectum.

The given parabola is x2 = 12y.

On comparing this equation with x2 = 4ay, we obtain 4a = 12 a = 3

The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x2 = 12 (3) x2 = 36 x = 6


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The coordinates of A are (6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of OAB are O (0, 0), A (6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit2.

Question 7:

A man running a racecourse notes that the sum of the distances from the two flag posts

form him is always 10 m and the distance between the flag posts is 8 m. find the equation

of the posts traced by the man.

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from

two fixed points is constant, then the path is an ellipse and this constant value is equal to

the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis

is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking themajor axis along the x-axis, the ellipse can be diagrammatically represented as


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The equation of the ellipse will be of the form , where a is the semi-major axis

Accordingly, 2a = 10 a = 5

Distance between the foci (2c) = 8

c = 4

On using the relation , we obtain

Thus, the equation of the path traced by the man is

Question 8:

An equilateral triangle is inscribed in the parabola y2

= 4 ax, where one vertex is at thevertex of the parabola. Find the length of the side of the triangle.

Let OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x-axis at point C.


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Let OC = k

From the equation of the given parabola, we have

The respective coordinates of points A and B are

AB = CA + CB =

Since OAB is an equilateral triangle, OA2 = AB2.

Thus, the side of the equilateral triangle inscribed in parabola y2 = 4 ax is .

3 dimensional geometry

Question 1:

A point is on the x-axis. What are its y-coordinates and z-coordinates?

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.


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Question 2:

A point is in the XZ-plane. What can you say about its y-coordinate?

If a point is in the XZ plane, then its y-coordinate is zero.

Question 3:

Name the octants in which the following points lie:

(1, 2, 3), (4, 2, 3), (4, 2, 5), (4, 2, 5), (4, 2, 5), (4, 2, 5),

(3, 1, 6), (2, 4, 7)

The x-coordina

Xi Coordinate

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