Question

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

• A. 89x - 52 y - 162 = 0
• B. 52x + 89y + 519 = 0
• C. 4x - y - 15 = 0
• D. 4x + 3y + 13 = 0

Answer 1

Answer: (B)

Given, equation

$4x+y-1=0\dots$ (i)
and $3x-4y+1=0\dots$ (ii)
Slope of line (i) $(m_1)=-4$
and slope of line (ii) $m_2=\frac{3}{4}$
Here, $AB=AC$
$\therefore △ABC$ is an isosceles.
$AB$ and $AC$ both passes through points $(2,-7)$.
Now, $\left(m_3\right)\tan \theta =\left|\frac{-4-\frac{3}{4}}{1+(-4)\left(\frac{3}{4}\right)}\right|$
$=\left|\frac{-\frac{19}{4}}{-2}\right|=\left|-\frac{19}{-8}\right|=\frac{19}{8}$
Now, required equation of line $AC$ is
$(y+7)=\frac{\frac{3}{4}-\frac{19}{8}}{1+\left(\frac{3}{4}\right)\left(\frac{19}{8}\right)}(x-2)$
$\Rightarrow (y+7)=\frac{\left(\frac{24-76}{32}\right)}{\left(\frac{32+57}{32}\right)}(x-2)$
$\Rightarrow y+7=-\frac{52}{89}(x-2)$
$\Rightarrow 89y+623=-52x+104$
$\Rightarrow 52x+89y+519=0$