Question

A straight line through origin $O$ meets the lines $3y=10-4x$ and $8x+6y+5=0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio:

• A. $4:1$
• B. $3:4$
• C. $2:3$
• D. $1:2$

Answer 1

The correct option is A $4:1$


We take the line $AB$ through origin which meets the two parallel lines.
$3y=10-4x$
$\Rightarrow y=\frac{-4}{3}x+\frac{10}{3}$
$\therefore OP=\frac{10}{3}$
$8x+6y+5=0$
$\Rightarrow y=\frac{-4}{3}x-\frac{5}{6}$
$\therefore OQ=\frac{5}{6}$
We observe that $△OAP\sim △OBQ$
$\therefore \frac{OA}{OB}=\frac{OP}{OQ}$
$\Rightarrow \frac{OA}{OB}=4$