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, the point O divides the segment PQ in the ratio

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, the point O divides the segment PQ in the ratio (A) 1:2 (B) 3: 4 (C) 2:1 (D) 4: 3

Tardigrade Admin · Accepted Answer

Now, distance of origin from 4x + 2y - 9 = 0 is (| -9 |/√42+22)=(9/√20) and distance of origin from 2x+ y + 6 = 0 is (| 6 |/√22+12)=(6/√5) Hence, the required ratio =(9/√20/6/√5)=(3/4)

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

$1 : 2$

$3 : 4$

$2 : 1$

$4 : 3$

Now, distance of origin from $4 x + 2 y - 9 = 0$ is
$\frac{∣ - 9∣}{4 ^{2} + 2 ^{2}} = \frac{9}{20}$
and distance of origin from $2 x + y + 6 = 0$ is
$\frac{∣6∣}{2 ^{2} + 1 ^{2}} = \frac{6}{5}$
Hence, the required ratio $= \frac{9/ 20}{6/ 5} = \frac{3}{4}$

Q. 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, the point O divides the segment PQ in the ratio

1:2

3:4

2:1

4:3