Question

If $(-4,6),(2,3)$ and $(-2,-5)$ are vertices of a triangle, then its incentre is

• A. $(-1,2)$
• B. $(2,-1)$
• C. $(1,2)$
• D. $(2,1)$

Answer 1

The correct option is A $(-1,2)$

By distance formula

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$a=\sqrt{(2+2{)}^2+(3+5{)}^2}=\sqrt{80}=4\sqrt{5}$ units

$b=\sqrt{(-2+4{)}^2+(-5-6{)}^2}=\sqrt{125}=5\sqrt{5}$ units

$c=\sqrt{(2+4{)}^2+(3-6{)}^2}=\sqrt{45}=3\sqrt{5}$ units

Inunbe Formula,

$x_c=\frac{a{x}_1+b{x}_1+c{x}_3}{a+b+c}$

$=\frac{4\sqrt{5}(-4)+5\sqrt{5}\left(2\right)+3\sqrt{5}(-2)}{4\sqrt{5}+5\sqrt{5}+3\sqrt{5}}=-1$

$y_c=\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}$

$=\frac{4\sqrt{5}\left(6\right)+5\sqrt{5}\left(3\right)+3\sqrt{5}(-5)}{4\sqrt{5}+\sqrt{5}+3\sqrt{5}}=2$

Inunbe $\equiv (x_c,y_c)\equiv (-1,2)$