Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-5) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 6.28

Ques.2. Find the points of trisection of the line segment joining the points:
(a) 5, −6 and (−7, 5),
(b) (3, −2) and (−3, −4),
(c) (2, −2) and (−7, 4).
Ans. The co-ordinates of a point which divided two points (x₁,y₁) and (x₂,y₂) internally in the ratio m:n is given by the formula,

The points of trisection of a line are the points which divide the line into the ratio 1:2.
(i) Here we are asked to find the points of trisection of the line segment joining the points A(5,−6) and B(−7,5). So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.
Let P(x,y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.

Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are.
(ii) Here we are asked to find the points of trisection of the line segment joining the points A(3,−2) and B(−3,−4).
So we need to find the points which divide the line joining these two points in the ratio 1 : 2 and 2 : 1.
Let P(x, y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.

Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are.
(iii) Here we are asked to find the points of trisection of the line segment joining the points A(2,−2) and B(−7,4).
So we need to find the points which divide the line joining these two points in the ratio 1 : 2 and 2 : 1.
Let P(x, y) be the point which divides the line joining ‘AB’ in the ratio 1 : 2.
(x,y) = (-1,0)
Let Q(e, d) be the point which divides the line joining ‘AB’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are (-1,0) and (-4,2).

Page No 6.28

Ques.3. Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (−2,−1), (1, 0), (4, 3) and(1, 2) meet.
Ans. The co-ordinates of the midpoint (x_m,y_m) between two points (x₁,y₁) and (x₂,y₂) is given by,

In a parallelogram the diagonals bisect each other. That is the point of intersection of the diagonals is the midpoint of either of the diagonals. Here, it is given that the vertices of a parallelogram are A(−2,−1), B(1,0) and C(4,3) and D(1,2).
We see that ‘AC’ and ‘BD’ are the diagonals of the parallelogram.
The midpoint of either one of these diagonals will give us the point of intersection of the diagonals.
Let this point be M(x, y).
Let us find the midpoint of the diagonal ‘AC’.

Hence the co-ordinates of the point of intersection of the diagonals of the given parallelogram are (1,1).

Page No 6.28

Ques.4. Prove that the points (3,−2), (4,0), (6,−3) and (5,−5) are the vertices of a parallelogram.
Ans. Let A (3,−2); B (4, 0); C (6,−3) and D (5,−5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram. We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.
Now to find the mid-point P(x,y) of two points A(x₁,y₁) and B(x₂,y₂) we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.
Hence ABCD is a parallelogram.

Page No 6.28

Ques.5. If P(9a −2,−b) divides the line segment joining A (3a+1,-3) and B (8a, 5) in the ratio 3 : 1 ,find the values of a and b.
Ans. It is given that P divides AB in the ratio 3 : 1.
Therefore, by section formula we have
⇒9a−2 = (3(8a)+1(3a+1))/2
⇒4(9a−2)=24a+3a+1
⇒36a−8=27a+1
⇒9a=9
⇒a=1
And,
⇒−b=(3(5)+1(−3))/3+1
⇒−4b=15−3
⇒b=−3

Page No 6.28

Ques.6. If (a,b) is the mid-point of the line segment joining the points A (10, −-6) , B (k,4) and a−-2b = 18 , find the value of k and the distance AB.
Ans. It is given that A(10, −6) and B(k, 4).
Suppose (a, b)  be midpoint of AB. Then,
a=(10+k)/2, b=(−6+4)/2=−2/2=−1
Now, a−2b=18
⇒a=18−2=16
Therefore,
16×2=10+k
⇒k=22
Further,

Page No 6.29.

Ques.7. Find the ratio in which the point (2, y) divides the line segment joining the points A (−2, 2) and B (3, 7). Also, find the value of y.
Ans. The co-ordinates of a point which divided two points  and  internally in the ratio  is given by the formula,

Here we are given that the point P(2,y) divides the line joining the points A(−2,2) and B(3,7) in some ratio.
Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

2m+2n = 3m-2n
m=4n
m/n = 4/1
We see that the ratio in which the given point divides the line segment is 4:1.
Let us now use this ratio to find out the value of ‘y’.

Equating the individual components we have

y=6
Thus the value of ‘y’ is 6.

Page No 6.29

Ques.8. If A (−1, 3), B (1, −1) and C (5, 1) are the vertices of a triangle ABC, find the length of the median through A.
Ans. The distance d between two points (x₁,y₁) and (x₂,y₂) is given by the formula

The co-ordinates of the midpoint (x_m,y_m) between two points (x₁,y₁) and (x₂,y₂) is given by,

Here, it is given that the three vertices of a triangle are A(−1,3), B(1,−1) and C(5,1).
The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.
Let ‘D’ be the mid-point of the side ‘BC’.
Let us now find its co-ordinates.

(x_D,y_D) = (3,0)
Thus we have the co-ordinates of the point as D(3,0).
Now, let us find the length of the median ‘AD’.

AD = 5
Thus the length of the median through the vertex ‘A’ of the given triangle is 5 units.

Page No 6.29

Ques.9. If the points P, Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B(7, 10) in 5 equal parts, find x, y and p.
Ans.
It is given that P, Q(x, 7), R, S(6, y) divides the line segment joining A(2, p) and B(7, 10) in 5 equal parts.
∴ AP = PQ = QR = RS = SB.....(1)
Now,
AP + PQ + QR + RS + SB = AB
⇒ SB + SB + SB + SB + SB = AB[From (1)]
⇒ 5SB = AB
⇒ SB = 1515AB.....(2)
Now,
AS = AP + PQ + QR + RS = 1/5 AB +1/5 AB + 1/5 AB + 1/5 AB = 4/5 AB.....(3)
From (2) and (3), we get
AS : SB = 4/5 AB : 1/5 AB = 4 : 1
Similarly,
AQ : QB = 2 : 3
Using section formula, we get
Coordinates of Q =
∴(x,7)=
⇒x=4 and 7=(20+3p)/5
Now,
7=(20+3p)/5
⇒20+3p=35
⇒3p=15
⇒p=5
Coordinates of S ==(6,9)
∴(6,y)=(6,9)⇒y=9
Thus, the values of x, y and p are 4, 9 and 5, respectively.

Page No 6.29

Ques.10. If a vertex of a triangle be (1, 1) and the middle points of the sides through it be (−2,−3) and (5 2) find the other vertices.
Ans. Let a ΔABC in which P and Q are the mid-points of sides AB and AC respectively. The coordinates are: A (1, 1); P (−2, 3) and Q (5, 2).
We have to find the co-ordinates of B(x₁,y₁) and C(x₂,y₂).
In general to find the mid-point P(x,y) of two points A(x₁,y₁) and B(x₂,y₂) and we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,
x₁=-5
y₁=5
So, co-ordinates of B is (−5, 5)
Similarly, mid-point Q of side AC can be written as,

Now equate the individual terms to get,
x₂=9
y₂=3
So, co-ordinates of C is (9, 3)

Page No 6.29

Quest.11. (i) In what ratio is the line segment joining the points (−2,−3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.
(ii) In what ratio is the line segment joining (−3, −1) and (−8, −9) divided at the point (−5, −21/5)?
Ans. (i) The ratio in which the y-axis divides two points (x₁,y₁) and (x₂,y₂) is λ:1
The co-ordinates of the point dividing two points (x₁,y₁) and (x₂,y₂) in the ratio m:n is given as,
;where λ=m/n
Here the two given points are A(−2,−3) and B(3,7).
Since, the point is on the y-axis so, x coordinate is 0.
(3λ−2)/1=0
⇒λ=2/3
Thus the given points are divided by the y-axis in the ratio 2:3.
The co-ordinates of this point (x, y) can be found by using the earlier mentioned formula.



(x,y) = (0,1)
Thus the co-ordinates of the point which divides the given points in the required ratio are (0,1).
(ii) The co-ordinates of a point which divided two points (x₁,y₁) and (x₂,y₂) internally in the ratio m:n is given by the formula,

Here it is said that the pointdivides the points (−3,−1) and (−8,−9). Substituting these values in the above formula we have,

Equating the individual components we have,

-5m-5n = -8m-3n
3m = 2n
m/n = 2/3
Therefore the ratio in which the line is divided is 2:3.

Page No 6.29

Ques.12. If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of k.
Ans. We have two points A (3, 4) and B (k, 7) such that its mid-point is P(x,y).
It is also given that point P lies on a line whose equation is
2x+2y+1=0
In general to find the mid-point P(x,y) of two points A(x₁,y₁) and B(x₂,y₂) we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,
x=(k+3)/2
y=11/2
Since, P lies on the given line. So,
2x+2y+1=0
Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,
k+15=0
So, k=-15

Page No 6.29

Ques.13. Find the ratio in which the point P(3/4,5/12) divides the line segment joining the points A(1/2,3/2) and B(2,−5)2,-5.    [CBSE 2015]
Ans. Suppose P(3/4,5/12) divides the line segment joining the points A(1/2,3/2) and B(2,−5)2,-5 in the ratio k : 1.
Using section formula, we get
Coordinates of P =
∴
⇒
Now,

⇒8k+2=3k+3
⇒5k=1
⇒k=1/5
Putting k = 1/5 inwe get
LHS==5/12 = RHS
Thus, the required ratio is 1/5 : 1 or 1 : 5.

Page No 6.29

Ques.14. Find the ratio in which the line segment joining (−2, −3) and (5, 6) is divided by (i) x-axis (ii) y-axis. Also, find the coordinates of the point of division in each case.
Ans. The ratio in which the x−axis divides two points (x₁,y₁) and (x₂,y₂) is λ:1
The ratio in which the y-axis divides two points (x₁,y₁) and (x₂,y₂) is μ:1
The co-ordinates of the point dividing two points (x₁,y₁) and (x₂,y₂) in the ratio m:n is given as,

Where λ = m/n
Here the two given points are A(−2,−3) and B(5,6).
i. The ratio in which the x-axis divides these points is
(6λ−3)/3=0
λ=1/2
Let point P(x, y) divide the line joining ‘AB’ in the ratio 1:2
Substituting these values in the earlier mentioned formula we have,



Thus the ratio in which the x−axis divides the two given points and the co-ordinates of the point is .
ii. The ratio in which the y-axis divides these points is
(5μ−2)/3=0
⇒μ=2/5
Let point P(x, y) divide the line joining ‘AB’ in the ratio 2:5
Substituting these values in the earlier mentioned formula we have,



Thus the ratio in which the x-axis divides the two given points and the co-ordinates of the point is.

Page No 6.29

Ques.15. Prove that the points (4, 5) (7, 6), (6, 3) (3, 2) are the vertices of a parallelogram. Is it a rectangle.
Ans. Let A(4, 5); B(7, 6); C(6, 3) and D(3, 2) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.
We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.
Now to find the mid-point P(x,y) of two points A(x₁,y₁) and B(x₂,y₂) we use section formula as,

So the mid-point of the diagonal AC is,

= (5,4)
Similarly mid-point of diagonal BD is,

= (5,4)
Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.
Hence ABCD is a parallelogram.
Now to check if ABCD is a rectangle, we should check the diagonal length.

Similarly,

Diagonals are of different lengths.
Hence ABCD is not a rectangle.