Question

If the coordinates of in-centre of the triangle whose vertices are $(0,6),(8,12)$ and $(8,0)$ is $(m,6)$. Then, find the value of $m$.

Answer 1

Let $A(0,6),B(8,12)$ and $C(8,0)$ be the vertices of triangle ABC.
Then $c=AB=\sqrt{{(8-0)}^2+{(12-6)}^2}=\sqrt{64+36}=\sqrt{100}=10$
Also, $b=CA=\sqrt{{(8-0)}^2+{(0-6)}^2}=\sqrt{64+36}=\sqrt{100}=10$
And $a=BC=\sqrt{{(8-8)}^2+{(0-12)}^2}=\sqrt{0+144}=\sqrt{144}=12$
The coordinates of the in-centre are
$(\frac{a{x}_1+{bx}_2+{cx}_3}{a+b+c},\frac{a{y}_1+{by}_2+{cy}_3}{a+b+c})=(\frac{12\times 0+10\times 8+10\times 8}{12+10+10},\frac{12\times 6+10\times 12+10\times 0}{12+10+10})=(\frac{160}{32},\frac{192}{32})=(5,6)$