Question

A line $4x+y=1$ through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. If $AB=AC$, then the equation of the line $AC$ is

• A. $52x-89y+519=0$
• B. $52x+89y-519=0$
• C. $52x+89y+519=0$
• D. None of these

Answer 1

The correct option is C $52x+89y+519=0$

Since, $AB=AC$
$\therefore \angle ABC=\angle ACB=\theta$ (say)


Now, slope of $BC=\frac{3}{4}$ and slope of $AB=-4$
Let slope of $AC$ be $m$.
Equating the two values of $\tan \theta$, we get
$\left|\frac{-4-\frac{3}{4}}{1-4.\frac{3}{4}}\right|=\left|\frac{\frac{3}{4}-m}{1+\frac{3}{4}m}\right|\Rightarrow \pm \frac{19}{8}=\frac{3-4m}{4+3m}$
$\Rightarrow 76+57m=24-32m\Rightarrow m=-\frac{52}{89}$
or $76+57m=32m-24\Rightarrow m=-4(\text{not possible as it is slope of}AB$
$\therefore$ Equation of $AC$ is $(y+7)=-\frac{52}{89}(x-2)$
$\Rightarrow 52x+89y+519=0$