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

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