A straight line through the origin O meets the parallel lines 4x+2y =9 and 2x+y+6=0 at points P and Q - Maths - Straight Lines

Question

A straight line through the origin O meets the parallel lines
4x+2y =9 and 2x+y+6=0 at points P and Q respectively, then point O
divides the line segment pq in the ratio _

Ayush Jha · Accepted Answer

Let the equation of the line passing through origin and intersecting the two given parallel lines be y = mx.

So, let us find the point of intersection of the y=mx line with the the given two parallel lines in terms of 'm' in order to get the co-ordinates of points P and Q.

So,

$O n s o l v i n g : - y = m x y = \frac{9 - 4 x}{2} S o w e g e t c o - o r d i n a t e s o f P, x = \frac{9}{2 (m + 2)} y = \frac{9 m}{2 (m + 2)} S i m i l a r l y w i t h t h e o t h e r l i n e y = m x y = - (6 + 2 x) W e g e t c o - o r d i n a t e s o f Q x = - \frac{6}{2 + m} y = - \frac{6 m}{2 + m}$

Now let us assume that origin divides the PQ in the ratio $λ : 1$ .

So applying the section formula we get,

$(\frac{\frac{9 λ}{2 (m + 2)} - \frac{6}{2 + m}}{λ + 1}, \frac{\frac{9 m λ}{2 (m + 2)} - \frac{6 m}{2 + m}}{λ + 1}) = (0, 0) E Q u a t i n g t h e a b s i c c a o n b o t h t h e s i d e s \frac{\frac{9 λ}{2 (m + 2)} - \frac{6}{2 + m}}{λ + 1} = 0 S i n c e d e n o \min a t o r c a n n o t b e 0. \frac{9 λ}{2 (m + 2)} = \frac{6}{2 + m} λ = \frac{4}{3}$

A straight line through the origin O meets the parallel lines 4x+2y =9 and 2x+y+6=0 at points P and Q respectively, then point O divides the line segment pq in the ratio _