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