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In what ratio does the line $\mathrm{y}-\mathrm{x}+2=0$ cut the line joining $(3,-1)$ and $(8,9) ?$
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The correct answer is:
$2: 3$
Let the point of intersection divide the line segment joining points, $(3,-1)$ and $(8,9)$ in $\mathrm{k}: 1$ ratio then:
The point is $\left(\frac{8 \mathrm{k}+3}{\mathrm{k}+1}, \frac{9 \mathrm{k}-1}{\mathrm{k}+1}\right)$
Since this point lies on the line $\mathrm{y}-\mathrm{x}+2=0$
We have, $\frac{9 \mathrm{k}-1}{\mathrm{k}+1}-\frac{8 \mathrm{k}+3}{\mathrm{k}+1}+2=0$
$=\frac{9 \mathrm{k}-1-8 \mathrm{k}-3}{\mathrm{k}+1}+2=0=\frac{\mathrm{k}-4}{\mathrm{k}+1}+2=0$
$=\mathrm{k}-4+2 \mathrm{k}+2=0=3 \mathrm{k}-2=0$
$\mathrm{k}=\frac{2}{3}: 1$ i.e. $2: 3$
The point is $\left(\frac{8 \mathrm{k}+3}{\mathrm{k}+1}, \frac{9 \mathrm{k}-1}{\mathrm{k}+1}\right)$
Since this point lies on the line $\mathrm{y}-\mathrm{x}+2=0$
We have, $\frac{9 \mathrm{k}-1}{\mathrm{k}+1}-\frac{8 \mathrm{k}+3}{\mathrm{k}+1}+2=0$
$=\frac{9 \mathrm{k}-1-8 \mathrm{k}-3}{\mathrm{k}+1}+2=0=\frac{\mathrm{k}-4}{\mathrm{k}+1}+2=0$
$=\mathrm{k}-4+2 \mathrm{k}+2=0=3 \mathrm{k}-2=0$
$\mathrm{k}=\frac{2}{3}: 1$ i.e. $2: 3$
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