Question

The angular points of a triangle are $A (-1, -7), B(5, 1)$ and $C(1, 4)$. The equation of the bisector of the $\angle$ABC is

• A. x = 7y + 2
• B. 7y = x + 2
• C. y = 7x + 2
• D. 7x = y + 2

Answer 1

Answer: (B)

Here, $A B=\sqrt{(5+1)^{2}+(1+7)^{2}}$
$=\sqrt{36+64}=10$

$B C=\sqrt{\left(1-2^{2}+(4-1)^{2}\right.}=\sqrt{16+9}=5$
By angle bisector theorem,
$A P: C P=10: 5=2: 1$
$\therefore P\left(\frac{2 \times 1+1 \times(-1)}{2+1}, \frac{2 \times 4+1 \times(-7)}{2+1}\right)=P\left(\frac{1}{3}, \frac{1}{3}\right)$
Required equation of $B P$ is
$y-1 =\frac{\frac{1}{3}-1}{\frac{1}{3}-5}(x-5) $
$\Rightarrow y-1 =\frac{-2}{-14}(x-5) $
$\Rightarrow 7 y-7 =x-5 $
$\Rightarrow 7 y =x+2$