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

Answer 1

Answer: (C)

Let equation of line thourgh $0(0,0)$ is $\frac{x}{\cos \theta }=\frac{y}{\sin \theta }=r$ If this line meets $3y=10-4x$ at A then
$3r,\sin \theta =10-4r_1\cos \theta$
$r_1(3\sin \theta +4\cos \theta )=10\dots \dots$ (i)
Again the line meets $8x+6y+5=0$ at $B$
$\Rightarrow 8r_2\cos \theta +6r_2\sin \theta +5=0$
$\Rightarrow 2r_2(3\sin \theta +4\cos \theta )=-5\dots \dots \dots$ (ii)
by $\frac{1}{2}$
$\Rightarrow \frac{r_1}{2r_2}=\frac{10}{-5}$
$\Rightarrow \frac{r_1}{r_2}=-\frac{4}{1}=4$