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 $3x -4 y + 1 = 0 \dots$ (ii)
Slope of line (i) $(m_1) = -4$
and slope of line (ii) $m_2 = \frac{3}{4}$
Here, $AB = AC$
$\therefore \triangle ABC$ 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 $A C$ is
$(y+7)=\frac{\frac{3}{4}-\frac{19}{8}}{1+\left(\frac{3}{4}\right)\left(\frac{19}{8}\right)}(x-2)$
$\Rightarrow (y+7)=\frac{\left(\frac{24-76}{32}\right)}{\left(\frac{32+57}{32}\right)}(x-2)$
$\Rightarrow y+7=-\frac{52}{89}(x-2)$
$\Rightarrow 89 y+623=-52 x+104$
$\Rightarrow 52 x+89 y+519=0$