If the orthocentre of the triangle formed by the lines $y=x+1, y=4 x-8$ and $y=m x+c$ is at $(3,-1)$, then $\mathrm{m}-\mathrm{c}$ is :

Question

Aicharya · Accepted Answer

Solve line PQ \& QR Point $\mathrm{Q}\left(\frac{1-\mathrm{c}}{\mathrm{m}-1}, \frac{1-\mathrm{c}}{\mathrm{m}-1}+1\right)$
$m_{2 H}=\frac{\frac{1-c}{m-1}+2}{\frac{1-c}{m-1}-3}=\frac{1-c+2 m-2}{1-c-3 m+3}=-\frac{1}{4}$
$\mathrm{m}_{2 \mathrm{H}}=\frac{\frac{1-\mathrm{c}}{\mathrm{m}-1}+2}{\frac{1-\mathrm{c}}{\mathrm{m}-1}-3}=\frac{1-\mathrm{c}+2 \mathrm{~m}-2}{1-\mathrm{c}-3 \mathrm{~m}+3}=-\frac{1}{4}$...(1)
$\because \mathrm{m}_{\mathrm{PH}}=\frac{5}{0} \rightarrow \infty$
$\Rightarrow$ Slope of line QR (m) $=0$
Put value of $m$ in equation (1)
$\frac{1-c-2}{1-c+3}=-\frac{1}{4} \Rightarrow c=0$
so $\mathrm{m}-\mathrm{c}=0$ Ans. ^

If the orthocentre of the triangle formed by the lines $y=x+1, y=4 x-8$ and $y=m x+c$ is at $(3,-1)$, then…

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