Question

If the vertices of a triangle are $A(2,4),B(2,6)$ and $C(2+\sqrt{3},5)$, then which of the following is/are true?

• A. Coordinates of the excentre opposite to vertex $A$ is $(2+\sqrt{3},7)$
• B. Coordinates of the excentre opposite to vertex $B$ is $(2-\sqrt{3},3)$
• C. Coordinates of the excentre opposite to vertex $C$ is $(2-\sqrt{3},5)$
• D. The area of $△ABC$ is $\sqrt{3}\text{sq. units}$

Answer 1

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

The area of $△ABC$ is $\sqrt{3}\text{sq. units}$

Given vertices of the triangle are $A(2,4),B(2,6)$ and $C(2+\sqrt{3},5)$
Now, finding the side lengths of the triangle,
$a=BC=\sqrt{(\sqrt{3}{)}^2+1^2}=2\text{units}$
$b=AC=\sqrt{(\sqrt{3}{)}^2+1^2}=2\text{units}$
$c=AB=\sqrt{0^2+2^2}=2\text{units}$

So, $△ABC$ is an equilateral.
Now, we know that
Excentre opposite to vertex $A$
$=(\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c},\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c})$
$=(-x_1+x_2+x_3,-y_1+y_2+y_3)$
$=(2+\sqrt{3},7)$

Excentre opposite to vertex $B$
$=(\frac{a{x}_1-b{x}_2+c{x}_3}{a-b+c},\frac{a{y}_1-b{y}_2+c{y}_3}{a-b+c})$
$=(x_1-x_2+x_3,y_1-y_2+y_3)$
$=(2+\sqrt{3},3)$

Excentre opposite to vertex $C$
$=(\frac{a{x}_1+b{x}_2-c{x}_3}{a+b-c},\frac{a{y}_1+b{y}_2-c{y}_3}{a+b-c})$
$=(x_1+x_2-x_3,y_1+y_2-y_3)$
$=(2-\sqrt{3},5)$

Now, area of an equilateral triangle
$=\frac{\sqrt{3}{a}^2}{4}=\frac{\sqrt{3}\times 2^2}{4}=\sqrt{3}\text{sq. units}$