Question

The incentre of the triangle formed by the vertices $A(-36,7),$ $B(20,7)$ and $C(0,-8)$ is

• A. $(\frac{1}{2},1)$
• B. $(0,-1)$
• C. $(1,\frac{1}{2})$
• D. $(-1,0)$

Answer 1

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

Assuming the vertices be
$A(x_1,y_1)=(-36,7)$
$B(x_2,y_2)=(20,7)$
$C(x_3,y_3)=(0,-8)$
Now, the side lengths are
$a=BC=25$ units
$b=AC=39$ units
$c=AB=56$ units
Now, the incentre,
$I=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
$\Rightarrow I=(\frac{25(-36)+39\left(20\right)+56\left(0\right)}{25+39+56},\frac{25\left(7\right)+39\left(7\right)+56(-8)}{25+39+56})\Rightarrow I=(\frac{-25\left(36\right)+39\left(20\right)}{120},\frac{64\left(7\right)-56\left(8\right)}{120})\Rightarrow I=(\frac{20(-45+39)}{120},\frac{64(7-7)}{120})$
$\therefore I=(-1,0)$