Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-8) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 6.30

Ques.45. Three vertices of a parallelogram are (a+b, a−b), (2a+b, 2a−b), (a−b, a+b). Find the fourth vertex.
Ans. Let ABCD be a parallelogram in which the co-ordinates of the vertices are;and. We have to find the co-ordinates of the forth vertex.
Let the forth vertex be D(x,y)
Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. 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,

The mid-point of the diagonals of the parallelogram will coincide.
So,
Coordinate of mid-point of AC = Co-ordinate of mid-point of BD
Therefore,

Now equate the individual terms to get the unknown value. So,
x=-b
y=b
So the forth vertex is D(-b,b).

Page No 6.30

Ques.46. If two vertices of a parallelogram are (3, 2) (−1, 0) and the diagonals cut at (2, −5), find the other vertices of the parallelogram.
Ans. We have a parallelogram ABCD in which A (3, 2) and B (−1, 0) and the co-ordinate of the intersection of diagonals is M (2,−5).
We have to find the co-ordinates of vertices C and D.
So let the co-ordinates be C(x₁,y₁) and D(x₂,y₂)
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,

The mid-point of the diagonals of the parallelogram will coincide.
So,
Co-ordinate of mid-point of AC = Co-ordinate of M
Therefore,

Now equate the individual terms to get the unknown value. So,
x = 1
y = -12
So the co-ordinate of vertex C is (1,−12)
Similarly,
Co-ordinate of mid-point of BD = Co-ordinate of M
Therefore,

Now equate the individual terms to get the unknown value. So,
x= 5
y= -10
So the co-ordinate of vertex C is (5,−10)

Page No 6.30

Ques.47. If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.
Ans. The co-ordinates of the midpoint (x_m,y_m) between two points (x₁,y₁) and (x₂,y₂) is given by,

Let the three vertices of the triangle be A(x_A, y_A), B(x_B,y_B) and C(x_C,y_C).
The three midpoints are given. Let these points be M_AB(3,4), M_BC(4,6) and M_CA(5,7).
Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,
x_A + x_B = 6
y_A + y_B = 8
Using the midpoint of another side we have,

Equating the individual components we get,
x_B + x_C = 8
y_B + y_C = 12
Using the midpoint of the last side we have,

Equating the individual components we get,
x_A + x_C = 10
y_A + y_C = 14
Adding up all the three equations which have variable ‘x’ alone we have,
x_A + x_B + x_B + x_C + x_A + x_C = 6 + 8 + 10
2(x_A + x_B + x_C) = 24
x_A + x_B + x_C = 12
Substituting x_B + x_C = 4 in the above equation we have,
x_A + x_B + x_C = 12
x_A + 8 = 12
_xA = 4
Therefore,
x_A + x_C = 10
x_{C = 10 - 4}
x_{C = 6}
And
x_A + x_B = 6
x_B = 6 - 4
xB = 2
Adding up all the three equations which have variable ‘y’ alone we have,
y_A + y_B + y_B + y_C + y_A + y_C = 8 + 12 + 14
2(y_A + y_B + y_C) = 34
y_A + y_B + y_C = 17
Substituting y_B + y_C = 12 in the above equation we have,
y_A + y_B + y_C = 17
y_A + 12 = 17
y_A = 5
Therefore,
y_A + y_C = 14
y_C = 14-5
y_C = 9
And
y_A + y_B = 8
y_B = 8-5
y_B = 3
Therefore the co-ordinates of the three vertices of the triangle are .

Page No 6.30

Ques.48. The line segment joining the points P(3, 3) and Q(6, −6) is trisected at the points A and B such that A is nearer to P. If A also lies on the line given by 2x + y + k = 0, find the value of k.
Ans. We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.
Let the co-ordinate of A be A(x₁,y₁)
Now according to the section formula if any point P divides a line segment joining A(x₁,y₁) and B(x₂,y₂) in the ratio m: n internally than,

The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2
Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point A is(4, 0)
It is given that point A lies on the line whose equation is
2x+y+k=0
So point A will satisfy this equation.
2(4)+0+k=0
So,
k=-8

Page No 6.30

Ques.49. If three consecutive vertices of a parallelogram are (1, −2), (3, 6) and (5, 10), find its fourth vertex.
Ans.

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1,−2);
B (3, 6) and C(5, 10). We have to find the co-ordinates of the forth vertex.
Let the forth vertex be D(x,y)
Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.
Now to find the mid-point P(x,y) of two points A(x₁,y₁) and B(x₂,y₂) we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.
So,
Co-ordinate of mid-point of AC = Co-ordinate of mid-point of BD
Therefore,

Now equate the individual terms to get the unknown value. So,

x=3
Similarly,

y=2
So the forth vertex is D(3,2)

Page No 6.30

Ques.50.
(i)  If the points A (a, −11), B (5, b), C(2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.
(ii) Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.
Ans. (i) Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (a,−11); B (5, b); C (2, 15) and D (1, 1).
Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.
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,

The mid-point of the diagonals of the parallelogram will coincide.
So,
Co-ordinate of mid-point of AC = Co-ordinate of mid-point of BD
Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

b=3
Therefore,
a=4, b=3
(ii) Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.
Diagonals of a parallelogram bisect each other.
∴ Mid point of AC = Mid point of BD
Mid point of (x₁, y₁) and (x₂, y₂) is .
Mid point of AC=
Mid point of BD=
=(9/2, 4/2)=(9/2, 2)
∴
⇒= 9/2 and  =2
⇒3+a=9 and 1+b=4
⇒a=6 and b=3
Hence,  the values of a and b is 6 and 3, respectively.

Page No 6.31

Ques.51. If the coordinates of the mid-points of the sides of a triangle be (3, −2), (−3, 1) and (4, −3), then find the coordinates of its vertices.
Ans. The co-ordinates of the midpoint (x_m,x_n) between two points (x₁,y₁) and (x₂,y₂) is given by,

Let the three vertices of the triangle be A(x_A,y_A), B(x_B,y_B) and C(x_C,y_C).
The three midpoints are given. Let these points be M_AB(3,-2), M_BC(-3,1) and M_CA(4,-3).
Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,
x_A + x_B = 6
y_A + y_B = -4
Using the midpoint of another side we have,

Equating the individual components we get,
x_B + x_C = -6
_yB + y_C = 2
Using the midpoint of the last side we have,

Equating the individual components we get,
x_A + x_C = 8
_yA + y_C = -6
Adding up all the three equations which have variable ‘x’ alone we have,
x_A + x_B + x_B + x_C + x_A + x_C = 6-6+8
2(x_A + x_B + x_C) = 8
x_A + x_B + x_C = 4
Substituting x_B + x_C = -6 in the above equation we have,
x_A + x_B + x_C = 4
x_A - 6 = 4
x_A = 10
Therefore,
x_A + x_C = 8
x_C = 8 - 10
x_C = -2
And
x_A + x_B = 6
x_B = 6 - 10
x_B = -4
Adding up all the three equations which have variable ‘y’ alone we have,
y_A + y_B + y_B + y_C + y_A + y_C = -4+2-6
2(y_A + y_B + y_C) = -8
y_A + y_B + y_C = -4
Substituting y_B + y_C = 2 in the above equation we have,
y_A + y_B + y_C = -4
y_A + 2 = -4
y_{A = -6}
Therefore,
y_A + y_C = -6
y_C = -6 + 6
y_C = 0
And
y_A + y_B = -4
y_B = -4 + 6
y_B = 2
Therefore the co-ordinates of the three vertices of the triangle are
.

Page No 6.31

Ques.52.
(i) Points P and Q trisect the line segment joining the points A(–2, 0) and B(0, 8) such that P is near to A. Find the coordinates of P and Q.
(ii) The line segment joining the points (3, −4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, −2) and (5/3, q) respectively. Find the values of p and q.
Ans. (ii) Let the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and Q(5/3, q).
Thus, AP = PQ = QB
Therefore, P divides AB internally in the ratio 1 : 2.
Section formula: if the point (x, y) divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m : n, then the coordinates (x, y) = (mx2+nx1m+n, my2+ny1m+n)mx2+nx1m+n, my2+ny1m+n

Therefore, using section formula, the coordinates of P are:

(p, −2)=(1(1)+2(3)1+2,1(2)+2(−4)1+2)⇒(p, −2)=(1+63,2−83)⇒(p, −2)=(73,−63)⇒(p, −2)=(73,−2)⇒p=73p, -2=11+231+2,12+2-41+2⇒p, -2=1+63,2-83⇒p, -2=73,-63⇒p, -2=73,-2⇒p=73

Also, Q divides AB internally in the ratio 2 : 1.

Therefore, using section formula, the coordinates of Q are:

(53, q)=(2(1)+1(3)2+1,2(2)+1(−4)2+1)⇒(53, q)=(2+33,4−43)⇒(53, q)=(53,03)⇒(53, q)=(53,0)⇒q=053, q=21+132+1,22+1-42+1⇒53, q=2+33,4-43⇒53, q=53,03⇒53, q=53,0⇒q=0

Hence,  the values of p and q is 7373 and 0, respectively.

Page No 6.31:

Question 53:

The line joining the points (2, 1) and (5, −8) is trisected at the points P and Q. If point P lies on the line 2x − y + k = 0. Find the value of k.

ANSWER:

We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point P is(3,−2)

It is given that point P lies on the line whose equation is

So point A will satisfy this equation.

So,

Page No 6.31:

Question 54:

A (4, 2), B(6, 5) and C (1, 4) are the vertices of ΔABC.

(i) The median from A meets BC in D. Find the coordinates of the point D.

(ii) Find the coordinates of point P and AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1

(iv) What do you observe?

ANSWER:

We have triangle in which the co-ordinates of the vertices are A (4, 2); B (6, 5) and C (1, 4)

(i)It is given that median from vertex A meets BC at D. So, D is the mid-point of side BC.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is 

(ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

P divides AD in the ratio 2: 1. So,

(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.

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

So co-ordinates of F is 

Similarly mid-point E of side AC can be written as,

So co-ordinates of E is 

Q divides BE in the ratio 2: 1. So,

Similarly, R divides CF in the ratio 2: 1. So,

(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.

Page No 6.31:

Question 55:

If the points A (6, 1), B (8, 2), C (9, 4) and D (k, p) are the vertices of a parallelogram taken in order, then find the values of k and p.

ANSWER:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (6, 1); B (8, 2); C (9, 4) and D (k, p).

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,