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
Answer :
3 : 4
Solution :
The given lines are
$$\eqalign{ & \,2x + y = \frac{9}{2}.....(1) \cr & {\text{and }}2x + y = - 6.....(2) \cr} $$
Signs of constants on R.H.S. show that two lines lie on opposite sides of origin. Let any line through origin meets these lines in $$P$$ and $$Q$$ respectively then required ratio is $$OP: OQ$$
$$\eqalign{ & {\text{Now in }}\Delta OPA\,\,{\text{and }}\Delta OQC, \cr & \angle POA = \angle QOC\,\,\left( {{\text{ver}}{\text{.}}\,\,{\text{opp}}{\text{.}}\,\,\angle 's} \right) \cr & \angle PAO = \angle OCQ\,\,\left( {{\text{alt}}{\text{. int}}{\text{.}}\,\,\angle 's} \right) \cr & \therefore \Delta OPA \sim \Delta OQC\,\,\left( {{\text{by }}AA{\text{ similarly}}} \right) \cr & \therefore \frac{{OP}}{{OQ}} = \frac{{OA}}{{OC}} = \frac{{\frac{9}{4}}}{3} = \frac{3}{4} \cr & \therefore \,{\text{Required ratio is 3 : 4}} \cr} $$
The given lines are
$$\eqalign{ & \,2x + y = \frac{9}{2}.....(1) \cr & {\text{and }}2x + y = - 6.....(2) \cr} $$
Signs of constants on R.H.S. show that two lines lie on opposite sides of origin. Let any line through origin meets these lines in $$P$$ and $$Q$$ respectively then required ratio is $$OP: OQ$$
$$\eqalign{ & {\text{Now in }}\Delta OPA\,\,{\text{and }}\Delta OQC, \cr & \angle POA = \angle QOC\,\,\left( {{\text{ver}}{\text{.}}\,\,{\text{opp}}{\text{.}}\,\,\angle 's} \right) \cr & \angle PAO = \angle OCQ\,\,\left( {{\text{alt}}{\text{. int}}{\text{.}}\,\,\angle 's} \right) \cr & \therefore \Delta OPA \sim \Delta OQC\,\,\left( {{\text{by }}AA{\text{ similarly}}} \right) \cr & \therefore \frac{{OP}}{{OQ}} = \frac{{OA}}{{OC}} = \frac{{\frac{9}{4}}}{3} = \frac{3}{4} \cr & \therefore \,{\text{Required ratio is 3 : 4}} \cr} $$