Question

P(x, y) is a point which divides line segment joining A(2, 1) and B(5, -8) in 1:2 ratio. Find

the point P. Hence find ‘k’ if point ‘P’ lies in the line 2x – y + k = 0.​

Answer 1

Answers :-

Co-ordinates of P are (3, -2)

Value of k is -8

Given :-

The point (x,y) divides the line segment A(2, 1) B(5,-8) in ratio 1:2

To find :-

The co-ordinates of P and the value of k for the equation 2x-y + k = 0

SOLUTION:-

Firstly we find the co-ordinates of point P by Section formula  

If a point P divides the point A (x1, y1) and Point B (x2,y2) is ratio m:n then the co-ordinates of point P are

$P \: = \bigg( \dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)$

So,

$(x_1,y_1 ) = (2,1) \\ (x_2,y_2) = (5, - 8)$

$m:n = 1:2$

Substituting the values,  

$(x ,\: y ) = \bigg( \dfrac{1(5)+ 2(2)}{1 + 2} , \dfrac{1( - 8)+ 2(1)}{1 + 2} \bigg)$

$(x ,\: y ) = \bigg( \dfrac{5+ 4}{3} , \dfrac{ - 8+ 2}{3} \bigg)$

$(x ,\: y ) = \bigg( \dfrac{9}{3} , \dfrac{ -6}{3} \bigg)$

$(x ,\: y ) = ( 3, - 2)$

So, the co-ordinates of P are (3,-2)

Now ,

Finding the value of k As they given line

2x -y + k =0

So, substituting the values,

• x =3
• y = -2

2(3) -(-2) + k = 0

6 + 2 + k = 0

8 + k = 0

k = -8

So, the value of k is -8