Question

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).

Answer 1

The coordinates of the in-centre of a triangle whose vertices are $A\left(x_1,y_1\right),B\left(x_2,y_2\right)\mathrm{and}C\left(x_3,y_3\right)$ are $\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$, where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
$a=BC=\sqrt{{\left(20-0\right)}^2+{\left(7+8\right)}^2}=25$
$b=AC=\sqrt{{\left(0+36\right)}^2+{\left(-8-7\right)}^2}=39$
$c=AB=\sqrt{{\left(20+36\right)}^2+{\left(7-7\right)}^2}=56$

Thus, the coordinates of the in-centre of the given triangle are:

$\left(\frac{25\times \left(-36\right)+39\times 20+0}{25+39+56},\frac{25\times 7+39\times 7+56\left(-8\right)}{25+39+56}\right)$

= $\left(\frac{-120}{120},0\right)$

= $\left(-1,0\right)$

Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is $\left(-1,0\right)$.