Question

If $A(-5,3),B(-3,3)$ and $C(-4,4)$ represents vertices of a triangle, then which of the following option(s) are correct?

• A. Incentre of $△ABC$ is $(-4,2+\sqrt{2})$
• B. Orthocentre centre of $△ABC$ is $(-4,4)$
• C. Excentre opposite to vertex $A$ is $(\sqrt{2}-4,4)$
• D. Excentre opposite to vertex $C$ is $(-4,2-\sqrt{2})$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Incentre of $△ABC$ is $(-4,2+\sqrt{2})$
B Orthocentre centre of $△ABC$ is $(-4,4)$
C Excentre opposite to vertex $A$ is $(\sqrt{2}-4,4)$
D Excentre opposite to vertex $C$ is $(-4,2-\sqrt{2})$

Given vertices of the triangle are $A(-5,3),B(-3,3)$ and $C(-4,4)$

Now, finding the sidelengths, we get
$a=BC=\sqrt{(1{)}^2+(-1{)}^2}=\sqrt{2}\text{units}b=AC=\sqrt{(-1{)}^2+(-1{)}^2}=\sqrt{2}\text{units}c=AB=\sqrt{(-2{)}^2+(0{)}^2}=2\text{units}$

So, $BC=AC$
By observation we get
$c^2+a^2+b^2$
$A{B}^2=B{C}^2+A{C}^2$
So $△ABC$ is a isosceles right angle triangle at $C$,
Now,
Orthocentre is same as coordinate of vertex $C$
$=(-4,4)$

$I=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

$=(\frac{-5\sqrt{2}-3\sqrt{2}-8}{2\sqrt{2}+2},\frac{3\sqrt{2}+3\sqrt{2}+8}{2\sqrt{2}+2})$

$=(-4,2+\sqrt{2})$

Excentre opposite to vertex $A$ $=(\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c},\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c})$

$=(\frac{5\sqrt{2}-3\sqrt{2}-8}{2},\frac{-3\sqrt{2}+3\sqrt{2}+8}{2})$
$=(\sqrt{2}-4,4)$

Excentre opposite to vertex $C$ $=(\frac{a{x}_1+b{x}_2-c{x}_3}{a+b-c},\frac{a{y}_1+b{y}_2-c{y}_3}{a+b-c})$

$=(\frac{-5\sqrt{2}-3\sqrt{2}+8}{2\sqrt{2}-2},\frac{3\sqrt{2}+3\sqrt{2}-8}{2\sqrt{2}-2})$

$=(-4,2-\sqrt{2})$