Question

The incentre of triangle whose vertices are $(-36,7)$, $(20,7)$ and $(0,-8)$ is :

• A. $(0,-1)$
• B. $(-1,0)$
• C. $(\frac{1}{2},1)$
• D. None of these

Answer 1

The correct option is B $(-1,0)$

Let $A=(-36,7)$ $B=(20,7)$ and $C=(0,-8)$

$\therefore a=BC=\sqrt{{(0-20)}^2+{(-8-7)}^2}=\sqrt{400+225}=\sqrt{625}=25$

$b=CA=\sqrt{{(-36-0)}^2+{(7+8)}^2}=\sqrt{1296+225}=\sqrt{1521}=39$

$c=AB=\sqrt{{(20+36)}^2+{(7-7)}^2}=\sqrt{3136+0}=\sqrt{3136}=56$

Let $I(x,y)$ be the in-centre of $△ABC$

$\therefore x=\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}$ and $y=\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}$

Substituting the above values in the above formula we get

$x=\frac{25\times -36+39\times 20+56\times 0}{25+39+56}=\frac{-120}{120}=-1$

and $y=\frac{25\times 7+39\times 7+56\times -8}{25+39+56}=\frac{0}{120}=0$

Thus, the in-centre is at $(-1,0)$