Question

The incentre of the triangle with vertices $(1,\sqrt{3}),(0,0)$ and $(2,0)$ is:

• A. $(1,\frac{\sqrt{3}}{2})$
• B. $(\frac{2}{3},\frac{1}{\sqrt{3}})$
• C. $(\frac{2}{3},\frac{\sqrt{3}}{2})$
• D. $(1,\frac{1}{\sqrt{3}})$

Answer 1

Answer: (D)

$(1,\frac{1}{\sqrt{3}})$

Let $A(1,\sqrt{3}),B(0,0)$ and $C(2,0)$ be the vertices of triangle ABC.
Then $c=AB=\sqrt{{(0-1)}^2+{(0-\sqrt{3})}^2}=\sqrt{1+3}=\sqrt{4}=2$
Also, $b=CA=\sqrt{{(2-1)}^2+{(0-\sqrt{3})}^2}=\sqrt{1+3}=\sqrt{4}=2$
And $a=BC=\sqrt{{(2-0)}^2+{(0-0)}^2}=\sqrt{4+0}=\sqrt{4}=2$
The coordinates of the in-centre are
$(\frac{a{x}_1+{bx}_2+{cx}_3}{a+b+c},\frac{a{y}_1+{by}_2+{cy}_3}{a+b+c})=(\frac{2\times 1+2\times 0+2\times 2}{2+2+2},\frac{2\times \sqrt{3}+2\times 0+2\times 0}{2+2+2})$

$=(\frac{6}{6},\frac{2\sqrt{3}}{6})=(1,\frac{\sqrt{3}}{2})$