RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16 B

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RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16 B

Question 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:
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:
Let P(x, y) and Q(p,q) be the points of trisection of the line segment. Joining A(2,1) and B(5, -8)
The, P(x, y) divide AB in the ratio 1 : 2.

By section formula, we have

Here m = 1 and n = 2

P(3, -2) is the 1st point of trisection of AB
Also Q(p, q) divides AB in the ratio 2 : 1
Here m = 2 and n = 1
(x1=2, y1=1) and (x2=5, y2=-8)

Q(4, -5) is the 2nd point of trisection of AB
Hence P(3, -2) and Q(4, -5) are the required point.

Question 4:
Let P(x, y) and Q(p, q) be the point of trisection of line segment A(-4, 0) and B(0, 6)
Then P(x, y) divides AB in the ratio 1 : 2

Here m = 1 and n = 2

Question 5:
Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.
Coordinates of P are:

Question 6:
Let (x, y) be the coordinates of a point P which divides the line joining A(4, -5) and B(4, 5) such that AP : AB = 2 : 5

Coordinates of P are (4, -1)

Question 7:
(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 8:
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 9:
C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)

Question 10:
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 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:
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:
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 (34 : 1) which is (3 : 4)

Question 14:
Let P(-6, a) divides the join of A(-3, -1) and B(-8, 9) in the ratio k : 1
Then the coordinates of P are given by

But, we are given P is (-6, a)

Hence, required ratio is 3 : 2 and a = 5

Question 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:
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:
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:
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 23 : 1 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:
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:
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:
Here 
Let G(x, y) be the centroid of ∆ABC, then

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

Question 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:
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:
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:
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:
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).

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