# R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 16 Coordinate Geometry

## Chapter 16 – Co-ordinate Geometry Exercise Ex. 16A

Question 1

Find the distance between the points:

(i) A(9,3) and B(15, 11)

(ii) A(7, -4) and B(-5, 1)

(iii) A(-6, -4) and B(9, -12)

(iv) A(1, -3) and B(4, -6)

(v) P(a + b, a – b) and Q(a – b, a + b)

(vi) P(a sin a, a cos a) and Q(a cos a, -a sin a)

Solution 1

(i) The given points are A(9,3) and B(15,11)

(ii) The given points are A(7,4) and B(-5,1)

(iii) The given points are A(-6, -4) and B(9,-12)

(iv) The given points are A(1, -3) and B(4, -6)

(v) The given points are P(a + b, a – b) and Q(a – b, a + b)

(vi) The given points are P(a sin a, a cos a) and Q(a cos a, – a sina)

Question 2

Find the distance of each of the following points from the origin:

(i) A(5, -12)

(ii) B(-5, 5)

(iii) C(-4, -6)

Solution 2

(i) The given point is A(5, -12) and let O(0,0) be the origin

(ii) The given point is B(-5, 5) and let O(0,0) be the origin

(iii) The given point is C(-4, -6) and let O(0,0) be the origin

Question 3

Find
all possible values of a for which the distance between the points A(a, -1)
and B(5, 3) is 5 units.

Solution 3

The
given points are A(a, -1) and B(5,3)

Question 4

Find all possible values of y for which
the distance between the points A(2, -3) and B(10,
y) is 10 units.

Solution 4

Question 5

Find the values of x for which the
distance between the points P(x, 4) and Q(9, 10) is
10 units.

Solution 5

Question 6

If the point A(x, 2) is equidistant from
the points B(8, -2) and C(2, -2), find the value of
x. Also, find the length of AB.

Solution 6

Question 7

If the point A(0,
2) is equidistant from the points B(3, p) and C(p,5),
find the value of p. Also, find the length of AB.

Solution 7

Question 8

Find the point on the x-axis which is equidistant from the points (-2, 5) and (-2, 9).

Solution 8

Let any point P on x – axis is (x,0) which is equidistant from A(-2, 5) and B(-2, 9)

Hence, there is no point on x – axis which is equidistant from A(-2, 5) and B(-2, 9)

Question 9

Find points on the x – axis, each o f which is at a distance of 10 units from the point A(11, -8).

Solution 9

Let A(11, -8) be the given point and let P(x,0) be the required point on x – axis

Then,

Hence, the required points are (17,0) and (5,0)

Question 10

Find the point on the y-axis which is
equidistant from the points A(6, 5) and B(-4, 3).

Solution 10

Question 11

If the point P(x, y) is equidistant from
the points A(5, 1) and B(-1, 5), prove that 3x = 2y.

Solution 11

Question 12

If P(x, y) is a point equidistant from the points A(6, -1) and B(2, 3). Show that x – y = 3

Solution 12

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get

Question 13

Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).

Solution 13

Let the required points be P(x,y), then

PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively

Hence, the point P is (3, -1)

Question 14

If the points A(4, 3) and B(x, 5) lie on
a circle with the centre O(2, 3), find the value of x.

Solution 14

Question 15

If the point C(-2,
3) is equidistant from the points A(3, -1) and B(x, 8), find the values of x.
Also, find the distance BC.

Solution 15

Question 16

If the point P(2,
2) is equidistant from the points A(-2, k) and B(-2k, – 3), find k. Also,
find the length of AP.

Solution 16

Question 17

If the point (x, y) is equidistant from
the points (a + b, b – a) and (a – b, a + b), prove that bx
= ay.

Solution 17

Question 18

Using the distance formula, show that the
given points are collinear.

(1, -1), (5, 2) and (9, 5)

Solution 18

Question 19

Using the distance formula, show that the
given points are collinear.

(6, 9), (0, 1) and (-6, -7)

Solution 19

Question 20

Using the distance formula, show that the
given points are collinear

(-1, -1), (2, 3) and (8, 11)

Solution 20

Question 21

Using the distance formula, show that the
given points are collinear.

(-2, 5), (0, 1) and (2, -3).

Solution 21

Question 22

Show that the points A(7,
10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

Solution 22

Question 23

Show that the points A(3,
0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right triangle.

Solution 23

Question 24

If A(5, 2), B(2, -2) and C(-2, t) are the
vertices of a right triangle with ∠B = 90°, then find the value of t.

Solution 24

Question 25

Prove that the points A(2,
4), B(2, 6) and C(2 +, 5) are the vertices of an equilateral triangle.

Solution 25

Question 26

Show that the points (-3, -3), (3, 3) and
(-3, 3) are the vertices of an equilateral triangle.

Solution 26

Question 27

Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of isosceles triangle. Calculate its area.

Solution 27

Let A(-5,6), B(3,0) and C(9,8) be the given points. Then

Question 28

Show that the points O(0, 0), A(3, ) and B(3, –) are the vertices of an equilateral triangle. Find the area of this triangle.

Solution 28

are the given points

Hence, DABC is equilateral and each of its sides being

Question 29

Show that the following points are the vertices of a square:

(i)A(3, 2), B(0, 5), C(-3, 2) and D(0, -1)

(ii)A(6, 2), B(2, 1), C(1, 5) and D(5, 6)

(iii)P(0, -2), Q(3, 1), R(0, 4) and S(-3, 1)

Solution 29

(i)The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)

Thus, all sides of quad. ABCD are equal and diagonals are also equal

(ii)Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD

Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad ABCD is a square.

(iii)Let P(0, -2), Q(3,1), R(0,4) and S(-3,1) be the angular points of quad. ABCD

Join PR and QSD

Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal

Hence, quad. PQRS is a square

Question 30

Show that the points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Solution 30

Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.

Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.

Hence, ABCD is a rhombus

Question 31

Show that the points A(3,
0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find its
area.

Solution 31

Question 32

Show that the points A(6,
1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its
area.

Solution 32

Question 33

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?

Solution 33

Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then

Diagonal AC  Diagonal BD

Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equal

Question 34

Show that A(1,
2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show
that ABCD is not a rectangle.

Solution 34

Question 35

Show that the following points are the vertices of a rectangle:

(i)A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3)

(ii)A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)

(i)A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)

Solution 35

(i) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle.

(ii)Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then

Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.

(iii)Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle

## Chapter 16 – Co-ordinate Geometry Exercise Ex. 16B

Question 1

Find the coordinates of the point which divides the join of A(-1, 7) and B(4, -3) in the ratio 2 : 3.

Solution 1

The end points of AB are A(-1,7) and B(4, -3)

Let the required point be P(x, y)

By section formula, we have

Hence the required point is P(1, 3)

Question 2

Find the coordinates of the points which divides the join of A(-5, 11) and B(4, -7) in the ratio 7 : 2.

Solution 2

The end points of PQ are P(-5, 11) and Q(4, -7_

By section formula, we have

Hence the required point is (2, -3)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Points P, Q, R and S divide the line
segment joining the points A(1, 2) and R(6, 7) in
five equal parts. Find the coordinates of P, Q and R.

Solution 5

Question 6

Points P, Q and R in that order are dividing a line segment joining A(1, 6) and B(5, -2) in four equal parts, find the coordinates of P, Q and R.

Solution 6

Points P, Q, R divide the line segment joining the points A(1,6) and B(5, -2) into four equal parts

Point P divide AB in the ratio 1 : 3 where A(1, 6), B(5, -2)

Therefore,  the point P is

Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)

Question 7

The line segment joining the points A(3, -4) and B(1, 2) is trisected at the points P(p, -2) and . Find the values of p and q.

Solution 7

Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.

Coordinates of P are:

Question 8

Find the coordinates of the midpoint of the line segment joining:

(i)A(3, 0) and B(-5, 4)

(ii)P(-11, -8) and Q(8, -2)

Solution 8

(i)The coordinates of mid – points of the line segment joining A(3, 0) and B(-5, 4) are

(ii)Let M(x, y) be the mid – point of AB, where A is (-11, -8) and B is (8, -2). Then,

Question 9

If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2, 11), find the value of p.

Solution 9

The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is

Also, given the midpoint of AB is (2, p)

p = 3

Question 10

The midpoint of the line segment A(2a, 4) and B(-2, 3b) is C(1, 2a + 1). Find the value of a and b.

Solution 10

C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)

Question 11

The line segment joining A(-2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

Solution 11

Let A(-2, 9) and B(6, 3) be the two points of the given diameter AB and let C(a, b) be the center of the circle

Then, clearly C is the midpoint of AB

By the midpoint formula of the co-ordinates,

Hence, the required point C(2, 6)

Question 12

Find
the coordinates of a point A, where AB is a diameter of a circle with centre
C(2, -3) and the other end of the diameter is B(1, 4).

Solution 12

A,
B are the end points of a diameter. Let the coordinates of A be (x, y)

The
point B is (1, 4)

The
center C(2, -3) is the midpoint of AB

The
point A is (3, -10)

Question 13

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)?

Solution 13

Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1

By section formula, the coordinates of p are

Hence, the required ratio of which is (3 : 4)

Question 14

Solution 14

Question 15

Find the ratio in which the point P(m, 6) divides the join of A(-4, 3) and B(2, 8). Also, find the value of m.

Solution 15

Let P divided the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1

the point P is

Question 16

Find the ratio in which the point (-3, k) divides the join of A(-5, -4) and B(-2, 3). Also, find the value of k.

Solution 16

Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1

Coordinates of point P

Question 17

In what ratio is the line segment joining A(2, -3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.

Solution 17

Let the x- axis cut the join of A(2, -3) and B(5, 6) in the ratio k : 1 at the point P

Then, by the section formula, the coordinates of P are

But P lies on the x axis so, its ordinate must be 0

So the required ratio is 1 : 2

Thus the x – axis divides AB in the ratio 1 : 2

Putting we get the point P as

Thus, P is (3, 0) and k = 1 : 2

Question 18

In what ratio is the line segment joining the points A(-2, -3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

Solution 18

Let the y – axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1

Then, by section formula, the co-ordinates of P are

But P lies on the y-axis so, its abscissa is 0

So the required ratio is which is 2 : 3

Putting we get the point P as

i.e., P(0, 1)

Hence the point of intersection of AB and the y – axis is P(0, 1) and P divides AB in the ratio 2 : 3

Question 19

In what ratio does the line x – y – 2 = 0 divide the line segment joining the points A(3, -1) and B(8, 9)?

Solution 19

Let the line segment joining A(3, -1) and B(8, 9) is divided by x – y – 2 = 0 in ratio k : 1 at p

Coordinates of P are

Thus the line x – y – 2 = 0 divides AB in the ratio 2 : 3

Question 20

Find the lengths of the medians of a ABC whose vertices are A(0, -1), B(2, 1) and C(0, 3).

Solution 20

Let D, E, F be the midpoint of the side BC, CA and AB respectively in ABC

Then, by the midpoint formula, we have

Hence the lengths of medians AD, BE and CF are given by

Question 21

Find the centroid of ABC whose vertices are A(-1, 0), B(5, -2) and C(8, 2).

Solution 21

Here

Let G(x, y) be the centroid of ABC, then

Hence the centroid of ABC is G(4, 0)

Question 22

If G(2, -1) is the centroid of a ABC and two of its vertices are A(1, -6) and B(-5,2), find the third vertex of the triangle.

Solution 22

Two vertices of ABC are A(1, -6) and B(-5, 2) let the third vertex be C(a, b)

Then, the co-ordinates of its centroid are

But given that the centroid is G(-2, 1)

Hence, the third vertex C of ABC is (-2, 7)

Question 23

Find the third vertex of a ABC if two of its vertices are B(-3, 1) and C(0, -2), and its centroid is at the origin.

Solution 23

Two vertices of ABC are B(-3, 1) and C(0, -2) and third vertex be A(a, b)

Then the coordinates of its centroid are

Hence the third vertices A of ABC is A(3, 1)

Question 24

Show that the points A(3, 1), B(0, -2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.

Solution 24

Let A(3,1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral

Join AC, BD. AC and BD, intersect other at the point O.

We know that the diagonals of a parallelogram bisect each other

Therefore, O is midpoint of AC as well as that of BD

Now midpoint of AC is

And midpoint of BD is

Mid point of AC is the same as midpoint of BD

Hence, A, B, C, D are the vertices of a parallelogram ABCD

Question 25

If the points P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.

Solution 25

Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.

Join the diagonals PR and SQ.

They intersect each other at the point O. We know that the diagonals of a parallelogram bisect each other.

Therefore, O is the midpoint of PR as well as that of SQ

Now, midpoint of PR is

And midpoint of SQ is

Hence the required values are a = 4 and b = 3

Question 26

If three consecutive vertices of a parallelogram ABCD are A(1, -2), B(3, 6) and C(5, 10), find the fourth vertex D.

Solution 26

Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD

Let D(a, b) be its fourth vertex. Join AC and BD.

Let AC and BD intersect at the point O.

We know that the diagonals of a parallelogram bisect each other.

So, O is the midpoint AC as well as that of BD

Midpoint of AC is

Midpoint of BD is

Hence the fourth vertices is D(3, 2)

Question 27

In what ratio does y-axis divide the line
segment joining the points (-4, 7) and (3, -7)?

Solution 27

Question 28

If the point Plies on the line segment joining the points A(3,
-5) 2 and B(-7, 9) then find the ratio in which P divides AB. Also, find the
value of y.

Solution 28

Question 29

Find the ratio in which the line segment
joining the points A(3, -3) and B(-2, 7) is divided
by x-axis. Also, find the point of division.

Solution 29

Question 30

The base QR of an equilateral triangle
PQR lies on x-axis. The coordinates of the point Q are (-4, 0) and origin is
the midpoint of the base. Find the coordinates of the points P and R.

Solution 30

Question 31

The base BC of an equilateral triangle
ABC lies on y-axis. The coordinates of point C are (0, -3). The origin is the
midpoint of the base. Find the coordinates of the points A and B. Also, find
the coordinates of anther point D such that ABCD is a rhombus.

Solution 31

Question 32

Find the ratio in which the points p(-1, y) lying on the line segment joining points A(-3, 10) and B(6, -8)
divides it. Also, find the value of y.

Solution 32

Question 33

ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P, Q, R and
S be the midpoints of AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Solution 33

Question 34

The midpoint P of the line segment
joining the points A(-10, 4) and B(-2, 0) lies on
the line segment joining the points C(-9, -4) and D(-4, y). Find the ratio in
which P divides CD. Also find the value of y.

Solution 34

## Chapter 16 – Co-ordinate Geometry Exercise Ex. 16C

Question 1

Find the area of ABC, whose vertices are:

(i) A(1, 2), B(-2, 3) and C(-3, -4)

(ii) A(-5, 7), B(-4, -5) and C(4, 5)

(iii) A(3, 8), B(-4, 2) and C(5, -1)

(iv) A(10, -6), B(2, 5) and C(-1, 3)

Solution 1

(i) Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices of the given ABC, then

(ii) The coordinates of vertices of ABC are A(-5, 7), B(-4, -5) and C(4, 5)

Here,

(iii) The coordinates of ABC are A(3, 8), B(-4, 2) and C(5, -1)

(iv) Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given PQR. Then,

Question 2

Find the area of quadrilateral ABCD whose
vertices are A(3, -1), B(9, -5), C(14, 0) and D(9 -19).

Solution 2

Question 3

Find the area of quadrilateral PQRS whose
vertices are P(-5, -3), Q(4, -6), R(2, -3) and S(1,
2).

Solution 3

Question 4

Find the area of quadrilateral ABCD whose
vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3,
4).

Solution 4

Question 5

Find the area of quadrilateral ABCD whose
vertices are A(-5, 7), B(-4, -5), C(-1, -6) and D(4,
5).

Solution 5

Question 6

Find the area of the triangle formed by
joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution 6

Question 7

A(7, -3), B(5, 3) and C(3, -1) are
the vertices of a ∆ABC and AD is its median. Prove that the median AD divides
∆ABC into two triangles of equal areas.

Solution 7

Question 8

Find the area of ∆ABC with A(1, -4) and midpoints of sides through A being (2, -1)
and (0, -1).

Solution 8

Question 9

A(6, 1), B(8, 2) and C(9, 4) are
the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the

Solution 9

Question 10

If the vertices of ∆ABC be A(1, -3), B(4,
p) and C(-9, 7) 15 square units, find the values of p.

Solution 10

Question 11

Find the value of k so that the area of
the triangle with vertices A(k + 1, 1), B(4, -3) and
C(7, -k) is 6 square units.

Solution 11

Question 12

For what value of k(k
> 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k +
1, 10) equal to 53 square units?

Solution 12

Question 13

Show that the following points are
collinear.

A(2, -2), B(-3, 8) and C(-1, 4)

Solution 13

Question 14

Show that the following points are
collinear.

A(-5, 1), B(5, 5) and C(10, 7)

Solution 14

Question 15

Show that the following points are collinear.

A(5, 1), B(1, -1) and C(11, 4)

Solution 15

Question 16

Show that the following points are
collinear.

A(8, 1), B(3, -4) and C(2, -5)

Solution 16

Question 17

Find the value of x for which the points
A(x, 2), B(-3, -4) and C(7, -5) are collinear.

Solution 17

Question 18

For what value of x are the points A(-3, 12), B(7, 6) and C(x, 9) collinear?

Solution 18

The given points are A(-3, 12), B(7, 6) and C(x, 9)

Question 19

For what value of y are the points P(1, 4), Q(3, y) and R(-3, 16) collinear?

Solution 19

Let P(1, 4), Q(3, y) and R(-3, 16)

Question 20

Find the value of y for which the points A(-3, 9), B(2, y) and C(4, -5) are collinear.

Solution 20

Question 21

For what values of k are the points A(8, 1), B(3, -2k) and C(k, -5) collinear.

Solution 21

Question 22

Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Solution 22

Vertices of ABC are A(2, 1), B(x, y) and C(7, 5)

The points A, B and C are collinear

area of ABC =0

Or 4x – 5y – 3 = 0

Question 23

Find a relation between x and y, if the
points A(x, y), B(-5,7) and C(-4, 5) are collinear.

Solution 23

Question 24

Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if .

Solution 24

The vertices of ABC are (a, 0), (0, b), C(1, 1)

The points A, B, C are collinear

Area of ABC = 0

ab – a – b = 0 a + b = ab

Dividing by ab

Question 25

If the points P(-3,
9), Q(a, b) and R(4, -5) are collinear and a + b = 1. find the values of a
and b.

Solution 25

Question 26

Find the area of ∆ABC with vertices A(0, -1), B(2, 1) and C(0, 3). Also, find the area of the triangle
formed by joining the midpoints of its sides. Show that the ratio of the
areas of two triangles is 4 :1.

Solution 26

## Chapter 16 – Co-ordinate Geometry Exercise Ex. 16D

Question 1

Points A(-1, y)
and B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.

Solution 1

Question 2

If the point A(0, 2) is equidistant from
the points B(3, p) and C(p, 5), find p.

Solution 2

Question 3

ABCD is a rectangle whose three vertices
are B(4, 0), C(4, 3) and D(0, 3). Find the length of
one of its diagonal.

Solution 3

Question 4

If the point P(k – 1, 2) is equidistant
from the points A(3, k) and B(k, 5), find the values of k.

Solution 4

Question 5

Find the ratio in which the point P(x, 2)
divides the join of A(12, 5) and B(4, -3).

Solution 5

Question 6

Prove that the diagonals of a rectangle
ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and
D(2, 6) are equal and bisect each other.

Solution 6

Question 7

Find the lengths of the medians AD and BE
of ∆ABC whose vertices are A(7, -3), B(5, 3) and
C(3, -1).

Solution 7

Question 8

If the point C(k, 4) divides the join of
A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.

Solution 8

Question 9

Find the point on x-axis which is
equidistant from points A(-1, 0) and B(5, 0).

Solution 9

Question 10

Find the distance between the points .

Solution 10

Distance between the points

Question 11

Find the value of a, so that the point
(3, a) lies on the line represented by 2x – 3y = 5.

Solution 11

Question 12

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

Solution 12

The points A(4,3) and B(x, 5) lie on the circle with center O(2,3)

OA and OB are radius of the circle.

Question 13

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Solution 13

The point P(x, y) is equidistant from the point A(7, 1) and B(3, 5)

Question 14

If the centroid of ABC having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a + b + c).

Solution 14

The vertices of ABC are (a, b), (b, c) and (c, a)

Centroid is

But centroid is (0, 0)

a + b + c = 0

Question 15

Find the centroid of ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).

Solution 15

The vertices of ABC are A(2, 2), B(-4, -4) and C(5, -8)

Centroid of ABC is given by

Question 16

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7 , 8)?

Solution 16

Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k : 1

The point C is

But C is (4, 5)

Thus, C divides AB in the ratio 2 : 3

Question 17

If the points A(2, 3), B(4, k)and C(6, -3) are collinear, find the value of k.

Solution 17

The points A(2, 3), B(4, k) and C(6, -3) are collinear if area of ABC is zero

But area of ABC = 0,

k = 0

## Chapter 16 – Co-ordinate Geometry Exercise MCQ

Question 1

The distance of the point P(-6,8) from the origin is

Solution 1

Question 2

The distance of the point (-3, 4) from x-axis is

(a) 3

(b) -3

(c) 4

(d) 5

Solution 2

Question 3

The point on x-axis which is equidistant from points

A(-1, 0) and B(5, 0) is

(a) (0, 2)

(b) (2, 0)

(c) (3, 0)

(d) (0, 3)

Solution 3

Question 4

If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) they y equals

(a) 5

(b) 7

(c) 12

(d) 6

Solution 4

Question 5

If the point C(k, 4) divides the join of the points A(2, 6) and B(5,1) in the ratio 2:3 then the value of k is

Solution 5

Question 6

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is

Solution 6

Correct option: (d)

Question 7

If A(1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a gm ABCD then the value of x is

Solution 7

Correct option: (b)

Question 8

If the points A(x, 2), B(-3, -4) and C(7, -5) are collinear then the value of x is

(a) -63

(b) 63

(c) 60

(d) -60

Solution 8

Question 9

The area of a triangle with vertices A(5, 0),  B(8, 0) and C(8,4) in square units is

(a) 20

(b) 12

(c) 6

(d) 16

Solution 9

Question 10

The area of ABC with vertices A(a, 0), O(0, 0) and

B(0, b) in square units is

Solution 10

Question 11

(a) -8

(b) 3

(c) -4

(d) 4

Solution 11

Question 12

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is

(a) 5

(b) 4

(c) 3

(d) 25

Solution 12

Correct option: (a)

Question 13

The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2:1 is

(a) (2, 4)

(b) (3, 5)

(c) (4, 2)

(d) (5, 3)

Solution 13

Question 14

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are

(a) (-6, 7)

(b) (6, -7)

(c) (4, 2)

(d) (5, 3)

Solution 14

Question 15

In the given figure P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and

B(1, -5). Then y equals

Solution 15

Question 16

The midpoint of segment AB is P(0, 4). If the coordinates of B are (-2, 3), then the coordinates of A are

(a) (2, 5)

(b) (-2, -5)

(c) (2, 9)

(d) (-2, 11)

Solution 16

Question 17

The point P which divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2:3 lies in the quadrant

(a) I

(b) II

(c) III

(d) IV

Solution 17

Question 18

If A(6, -7) and B(-1, -5) are two given points then the distance 2AB is

(a) 13

(b) 26

(c) 169

(d) 238

Solution 18

Question 19

Which point on x-axis is equidistant from the points

A(7, 6) and B(-3, 4)?

(a) (0, 4)

(b) (-4, 0)

(c) (3, 0)

(d) (0, 3)

Solution 19

Question 20

The distance of P(3, 4) from the x-axis is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 1 units

Solution 20

Question 21

In what ratio does the x-axis divide the join of A(2, -3) and B(5, 6)?

(a) 2:3

(b) 3:5

(c) 1:2

(d) 2:1

Solution 21

Question 22

In what ratio does the y-axis divide the join of P(-4, 2) and Q(8, 3)?

(a) 3:1

(b) 1:3

(c) 2:1

(d) 1:2

Solution 22

Question 23

If P(-1, 1) is the midpoint of the line segment joining

A(-3, b) and B(1, b + 4) then b =?

(a) 1

(b) -1

(c) 2

(d) 0

Solution 23

Question 24

The line 2x + y – 4 = 0 divides the line segment joining A(2, -2) and (3, 7) in the ratio

(a) 2:5

(b) 2:9

(c) 2:7

(d) 2:3

Solution 24

Question 25

If A(4, 2), B(6, 5) and C(1,4) be the vertices of ABC and AD is a median, then the coordinates of D are

Solution 25

Question 26

If A(-1, 0), B(5, -2) and C(8,2) are the vertices of a ABC then its centroid is

(a) (12, 0)

(b) (6, 0)

(c) (0, 6)

(d) (4, 0)

Solution 26

Question 27

Two vertices of ABC are A (-1, 4) and B(5, 2) and its centroid is G(0, -3). Then, the coordinates of are

(a) (4, 3)

(b) (4, 15)

(c) (-4, -15)

(d) (-15, -4)

Solution 27

Question 28

The points A(-4, 0), B(4, 0) and C(0,3) are the vertices of a triangle, which is

(a) isosceles

(b) equilateral

(c) scalene

(d) right-angled

Solution 28

Question 29

The point P(0, 6), Q(-5, 3) and R(3, 1)are the vertices of a triangle, which is

(a) equilateral

(b) isosceles

(c) scalene

(d) right-angled

Solution 29

Question 30

If the points A(2, 3), B(5, k) and C(6, 7) are collinear then

Solution 30

Question 31

If the point A (1, 2), O(0, 0) and C(a, b) are collinear then

(a) a = b

(b) a = 2b

(c) 2a = b

(d) a + b = 0

Solution 31

Question 32

The area of ABC with vertices A(3, 0), B(7, 0) and

C(8, 4) is

(a) 14 sq units

(b) 28 sq units

(c) 8 sq units

(d) 6 sq units

Solution 32

Question 33

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of each of its diagonals is

Solution 33

Question 34

If the distance between the points A(4, p) and B(1, 0) is 5 then

(a) p = 4 only

(b) p = -4 only

(c) p = ± 4

(d) p = 0

Solution 34

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