Question

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

• A. $52x+89y+519=0$
• B. $52x+89y-519=0$
• C. $82x+52y+519=0$
• D. $89x+52y-519=0$

Answer 1

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

Let $AB$ makes an angle $\alpha$ with $BC$

Since, $AB$$=$$AC$

$\Rightarrow \angle ABC=\angle ACB$

$\Rightarrow \tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{-4-\frac{3}{4}}{1+(-4)\left(\frac{3}{4}\right)}=\frac{19}{8}$

Let the gradient of $AC$ be $m$

$\therefore \tan \alpha =\left|\frac{m-\frac{3}{4}}{1+\frac{3m}{4}}\right|$ $\Rightarrow \frac{19}{8}=\pm \frac{4m-3}{4+3m}\Rightarrow m=-4$ or $-\frac{52}{89}$

But the gradient of $AB$ is $-4$

$\therefore$ Gradient of $AC$ is $=-\frac{52}{89}$

$\therefore$ Equation of line $AC$ is

$y+7=-\frac{52}{89}(x-2)$ $\Rightarrow 52x+89y+519=0$