Question

The vertices of a triangle are $A(0,-6),$ $B(-6,0)$ and $C(1,1)$ respectively. Then coordinates of the excentre opposite to vertex $A$ are

• A. $(-\frac{3}{2},-\frac{3}{2})$
• B. $(-4,\frac{3}{2})$
• C. $(\frac{3}{2},-\frac{3}{2})$
• D. $(-4,6)$

Answer 1

The correct option is D $(-4,6)$

$a=BC=\sqrt{(-6-1{)}^2+(0-1{)}^2}\Rightarrow a=\sqrt{50}=5\sqrt{2}$
$b=CA=\sqrt{(1-0{)}^2+(1+6{)}^2}\Rightarrow b=\sqrt{50}=5\sqrt{2}$
$c=AB=\sqrt{(0+6{)}^2+(-6-0{)}^2}\Rightarrow c=\sqrt{72}=6\sqrt{2}$

Coordinates of excentre opposite to vertex $A$
$x=\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c}\Rightarrow x=\frac{-5\sqrt{2}\cdot 0+5\sqrt{2}(-6)+6\sqrt{2}\left(1\right)}{-5\sqrt{2}+5\sqrt{2}+6\sqrt{2}}\Rightarrow x=\frac{-24\sqrt{2}}{6\sqrt{2}}=-4$

$y=\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c}\Rightarrow y=\frac{-5\sqrt{2}(-6)+5\sqrt{2}\cdot 0+6\sqrt{2}\left(1\right)}{-5\sqrt{2}+5\sqrt{2}+6\sqrt{2}}\Rightarrow y=\frac{36\sqrt{2}}{6\sqrt{2}}=6$

Hence, the coordinates of the excentre opposite to vertex $A$ are $(-4,6)$.