A line $4 x+y=1$ passes through the point $\mathrm{A}(2,-7)$ meets the line BC whose equation is $3 x-4 y+1=0$ at the point B . The equation of the line $A C$ so that $A B=A C$ is

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Aicharya · Accepted Answer

Slopes of $A B$ and $B C$ are -4 and $\frac{3}{4}$ respectively.

Let $\alpha$ be the angle between AB and BC . Then, $\tan \alpha=\left|\frac{-4-\frac{3}{4}}{1-4\left(\frac{3}{4}\right)}\right|=\frac{19}{8}$...(i) Since $A B=A C$ $\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{ACB}=\alpha$ $\therefore \quad$ the line $A C$ also makes an angle $\alpha$ with $B C$. If $m$ is the slope of the line $A C$, then its equation is $y+7=\mathrm{m}(x-2)$...(ii) Now, $\tan \alpha= \pm\left[\frac{\mathrm{m}-\frac{3}{4}}{1+\mathrm{m} \cdot \frac{3}{4}}\right]$ $\Rightarrow \frac{19}{8}= \pm \frac{4 \mathrm{~m}-3}{4+3 \mathrm{~m}}$ ...[From (i)] $\Rightarrow \mathrm{m}=-4$ or $-\frac{52}{89}$ But slope of $A B$ is -4 , so slope of AC is $-\frac{52}{89}$. Therefore, the equation of line AC given by (ii) is $52 x+89 y+519=0$.

A line $4 x+y=1$ passes through the point $\mathrm{A}(2,-7)$ meets the line BC whose equation is $3 x-4…

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