Question

Straight lines $x–y=7$ and $x+4y=2$ intersect at B. Points A and C are so chosen on these two lines such that AB = AC. The equation of line AC passing through (2, –7) is

• A. x – y – 9 = 0
• B. 23x + 7y + 3 = 0
• C. 2x – y – 11 = 0
• D. 7x – 6y – 56 = 0

Answer 1

The correct option is B 23x + 7y + 3 = 0

Let the slope of line AC be $m$
Then equation of AC be $\frac{y-(-7)}{x-2}=m$
Let the angle between line AB and BC be $\theta$
$\tan \theta =\left|\frac{1-\left(\frac{-1}{4}\right)}{1+1\cdot \left(\frac{-1}{4}\right)}\right|$
$\Rightarrow \tan \theta =\left|\frac{\frac{5}{4}}{\frac{3}{4}}\right|$
$\tan \theta =\frac{5}{3}$
Since AB=AC if we consider $△$ABC
$\angle$ABC=$\angle$ACB
Equating tangent of both angles we get
$\frac{5}{3}=\left|\frac{m-\left(\frac{-1}{4}\right)}{1+m\cdot \left(\frac{-1}{4}\right)}\right|$
$\Rightarrow \frac{5}{3}=\frac{m-\left(\frac{-1}{4}\right)}{1+m\cdot \left(\frac{-1}{4}\right)}$ or $\frac{5}{3}=-\left(\frac{m-\left(\frac{-1}{4}\right)}{1+m\cdot \left(\frac{-1}{4}\right)}\right)$
Solving we get
$m=1,\frac{-23}{7}$
m won't be 1 because we will get our AB line in that case.
So m=$\frac{-23}{7}$

$\therefore$ Equation of line is 23x + 7y + 3 = 0