A straight line through the origin O meets the parallel lines and at points and respectively. Then the point O divides the segment PQ in the ratio [2002]

Question

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 [2002]

Smart Achievers · Accepted Answer

(2)

The given lines are

$2 x + y = \frac{9}{2} \dots (i)$

$and 2 x + y = - 6 \dots (i i)$

Signs of constants on R.H.S. show that two lines lie on opposite sides of origin. Let a line through origin meets these lines in $P$ and $Q$ respectively. Then required ratio is $O P : O Q .$

$In ∆ O P A and ∆ O Q C,$

$∠ P O A = ∠ Q O C (vertically opposite angles)$

$∠ P A O = ∠ O C Q (alternate interior angles)$

$∴$ $∆ O P A ~ ∆ O Q C (AA similarity)$

$∴$ $\frac{O P}{O Q} = \frac{O A}{O C} = \frac{\frac{9}{4}}{3} = \frac{3}{4}$

$∴$ $Required ratio = 3 : 4$

Solution

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 [2002]

1 1 : 2

2 3 : 4

3 2 : 1

4 4 : 3

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