RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.2.

## Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2

Question 1.

Find the co-ordinates of the point which divides the line (RBSESolutions.com) segment joining the points (3, 5) and (7, 9) in the ratio 2 : 3 internally.

Solution :

Let point P(x, y) divides the line segment joining the point A(3, 5) and B(7, 9) internally in the ratio 2 : 3.

Hence, the coordinate of P is \(\left( 4\frac { 3 }{ 5 } ,6\frac { 3 }{ 5 } \right) \).

Question 2.

Find the co-ordinates of the point which (RBSESolutions.com) divides the line segment joining the points (5, -2) and \(\left( -1\frac { 1 }{ 2 } ,4 \right) \) in the ratio 7 : 9 externally.

Solution :

Let point P(x, y) divides the line segment joining the points A(5, -2) and B\(\left( -1\frac { 1 }{ 2 } ,4 \right) \) externaly in the ratio 7 : 9.

x_{1} = 5 y_{1} = -2 m_{1} = 7

x_{2} = -1\(\frac { 1 }{ 2 }\) y_{2} = 4 m_{2} = 9

By formula of external division

Hence, the coordinate of P is \(\left( 27\frac { 3 }{ 4 } ,-23 \right) \)

Question 3.

Prove that origin O divides the line joining (RBSESolutions.com) the points 4(1, -3) and B(-3, 9) in the ratio 1 : 3 internally. Find the co-ordinates of the points which externally divides the line.

Solution :

Let the point P(x, y) divides the line segment joining the point A(1, -3) and B(-3, 9) internally in the ratio 1 : 3.

Hence co-ordinate of P(x, y) is (0, 0). So origin O is divides (RBSESolutions.com) the line segment joining the given points internally in the ratio 1 : 3.

Again point P(x, y) divides externally, then by formula of external division

Hence, co-ordinate of point P which divides externally = (3, -9).

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Question 4.

Find the mid point of line joining (RBSESolutions.com) the points (22, 20) and (0, 16).

Solution :

Let co-ordinates of mid point is P (x, y) which ¡s line joining the points A(22, 20) and B(0, 16).

x_{1} = 22, y_{1} = 20, x_{2} = 0, y_{2} = 16

By formula of mid point

Hence, co-ordinate of mid point P(x, y) = (11, 18)

Question 5.

In which ratio, x-axis divides the line segment (RBSESolutions.com) which joins points (5, 3) and (-3, -2)?

Solution :

On x-axis, the ordinate of each point is zero. So let point P(x, 0) divides the line segment joining the points A(5, 3) and B(-3, -2) internally in the ratio m_{1} : m_{2}

Hence, line segment joining the given points by x-axis divides internally in the ratio 3 : 2.

Question 6.

In which ratio, y-axis, divides the line (RBSESolutions.com) segment which joins points (2, -3) and (5, 6)?

Solution :

On y-axis, coordinate of point x is zero, so let point P(0, y) divides the line segment joining the point A(2, -3) and B(5, 6) externally in the ratio m_{1} : m_{2}.

Then by formula of external division

⇒ 5m_{1} – 2m_{2} = 0

⇒ 5m_{1} = 2m_{2}

⇒ \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 2 }{ 5 } \)

⇒ m_{1} : m_{2} = 2 : 5

Hence, line segment joining these points divides externally in the ratio 2 : 5 by y-axis.

Question 7.

In which ratio, point (11, 15) divides the (RBSESolutions.com) line segment which joins (15, 5) and (9, 20)?

Solution :

Let the point P(1 1, 15) divides the line segment joining the point A(15, 5) and B(9, 20) internally in the ratio λ : 1

Hence x_{1} = 15 y_{1} = 5 m_{1} = λ

x_{2} = 9 y_{2} = 20 m_{2} = 1

⇒ 11(λ + 1) = 9λ + 15

⇒ 11λ + 11 = 9λ + 15

⇒ 11λ – 9λ = 15 – 11

⇒ 2λ = 4

λ = \(\frac { 4 }{ 2 }\) = \(\frac { 2 }{ 1 }\)

∴ λ : 1 = 2 : 1

Hence, required ratio is 2 : 1.

Question 8.

If point P (3, 5) divides line segment which (RBSESolutions.com) joins A(-2, 3) and B(x, y) in the ratio 4 : 7 internally, then find the co-ordinates of B.

Solution :

Let the coordinate of point B is (x, y)

Let the point P(3, 5) divides the line segment joining the points A(-2, 3) and B(x, y) internally in the inthe ratio 4 : 7.

So, x_{1} = -2, y_{1} = 3 m_{1} = 4

x_{2} = x y_{2} = y m_{2} = 7

By section formula

Hence, the co-ordinate of point B is \(\left( \frac { 47 }{ 4 } ,\frac { 17 }{ 2 } \right)\)

Question 9.

Find the co-ordinates of point which (RBSESolutions.com) trisects the line joining point (11, 9) and (1, 2).

Solution :

Let P(x_{1} y_{1}) and Q(x_{2}, y_{2}) are required points which is trisect the line segment joining the points A(11, 9) and B(1, 2) respectively.

Let AP = PQ = BQ = x

PB = x + x = 2x

AQ = x + x = 2x

Hence point P divides line segment AB in the (RBSESolutions.com) ratio 1 : 2 and point Q, divides the line segment Ab in the ratio 2 : 1.

By the section formula for point P,

m_{1} = 1, m_{2} = 2

∴ Hence the co-ordinate of point P is \(\left( \frac { 23 }{ 3 } ,\frac { 20 }{ 3 } \right)\)

for point Q,

m_{1} = 2, m_{2} = 1

So, the co-ordinate of Q = \(\left( \frac { 13 }{ 3 } ,\frac { 13 }{ 3 } \right)\)

Hence, required coordinate of point P and Q is \(\left( \frac { 23 }{ 3 } ,\frac { 20 }{ 3 } \right)\) and \(\left( \frac { 13 }{ 3 } ,\frac { 13 }{ 3 } \right)\) respectively.

Question 10.

Find the co-ordinates of point which quarter (RBSESolutions.com) sects the line joining point (-4, 0) and (0, 6).

Solution :

Let C, D and E are required point which quarter sects the line segment joining the points A(-4, 0) and B(0, 6)

Since D is mid point of A and B, C is is midpoint A and D, and E ¡s mid point of D and B, then AC = CD = DE = EB

Now, co-ordinate of mid point D of A and B.

= \(\left( \frac { -2 }{ 2 } ,\frac { 9 }{ 2 } \right)\)

= \(\left( -1,\frac { 9 }{ 2 } \right)\)

Hence, co-ordinate of points divided in four equal (RBSESolutions.com) part of AB are \(\left( -3,\frac { 3 }{ 2 } \right)\), (-2, 3) and \(\left( -1,\frac { 9 }{ 2 } \right)\) respectively.

Question 11.

Find the ratio in which line 3x + y = 9 divides the line segment which joins points (1, 3) and (2, 7)

Solution :

Let the line 3x + y = 9 divides the line segment joining A(1, 3) and B(2, 7) at point P(x, y) in the ratio λ : 1

x_{1} = 1 y _{1} = 3 m_{1} = λ

x_{2} = 2 y_{2} = 7 m_{2} = 1

⇒ 6λ + 3 + 7λ + 3 = 9(λ + 1)

⇒ 13λ + 6 = 9λ + 9

⇒ 13λ – 9λ = 9 – 6

⇒ 4λ = 3

⇒ λ = \(\frac { 3 }{ 4 }\)

λ : 1 = 3 : 4

Hence, required ratio is 3 : 4.

Question 12.

Find the ratio where point (-3, P), divides internally (RBSESolutions.com) the line segment which joins points (-5, -4) and (-2, 3). Also find P.

Solution :

Let the point P(-3, p) divides the line segment joining the points A(-5, -4) and B(-2, 3) internally in the ratio m_{1} : m_{2}.

Hence, x co-ordinate of point P(-3, p)

= -3(m_{1} + m_{2}) = -2m_{1} – 5m_{2}

⇒ -3m_{1} – 3m_{2} = -2m_{1} – 5m_{2}

⇒ -3m_{1} + 2m_{1} = -5m_{2} + 3m_{2}

⇒ – m_{1} = -2m_{2}

⇒ \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { -2 }{ -1 } \) = \(\frac { 2 }{ 1 }\)

Hence required ratio m_{1} : m_{2} = 2 : 1

Now, y co-ordinate of point P(-3, p)

Hence, required (RBSESolutions.com) ratio 2 : 1 and p = \(\frac { 2 }{ 3 }\)

We hope the given RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.2, drop a comment below and we will get back to you at the earliest.

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