Question

The incentre of the triangle formed by $(0,0),(5,12),(16,12)$ is:

• A. $(7,9)$
• B. $(9,7)$
• C. $(-9,7)$
• D. $(-7,9)$

Answer 1

The correct option is A $(7,9)$

Let $A(0,0),B(5,12)$ and $C(16,12)$ be the vertices of $△ABC.$
Then, $c=AB=\sqrt{{(5-0)}^2+{(12-0)}^2}=\sqrt{25+144}=\sqrt{169}=13$
Also, $b=CA=\sqrt{{(16-0)}^2+{(12-0)}^2}=\sqrt{256+144}=\sqrt{400}=20$
And $a=BC=\sqrt{{(16-5)}^2+{(12-12)}^2}=\sqrt{121+0}=\sqrt{121}=11$
The coordinates of the in-centre are $(\frac{a{x}_1+{bx}_2+{cx}_3}{a+b+c},\frac{a{y}_1+{by}_2+{cy}_3}{a+b+c})$

$=(\frac{11\times 0+20\times 5+13\times 16}{11+20+13},\frac{11\times 0+20\times 12+13\times 12}{11+20+13})$

$=(\frac{308}{44},\frac{396}{44})$

$=(7,9)$