Question

Find the incentre of the triangle
whose vertices are (2,4), 6, 4) and
(2.0).​

Answer 1

Given:

A triangle whose vertices are $(2,4), (6,4),$ and $(2,0)$.

To find:

Incentre of the triangle

Solution:

Let $A(2,4)$, $B(6,4)$ and $C(2,0)$

$a=BC=\sqrt{(2-6)^{2}+(0-4)^{2} } =\sqrt{16+16}=\sqrt{32}$

$a=BC=4\sqrt{2}$ sq. units

$b=CA=\sqrt{(2-2)^{2}+(0-4)^{2} } =\sqrt{16}$

b=CA=4 sq. units

$c=AB=\sqrt{(6-2)^{2}+(4-4)^{2} } =\sqrt{16}$

$c=AB=4$ sq. units

Incentre of triangle, $(x,y)=(\frac{ax_{1}+bx_{2}+cx_{3} }{a+b+c},\frac{ay_{1}+by_{2}+cy_{3} }{a+b+c} )$  

$(x,y)=(\frac{4\sqrt{2}(2)+4(6)+4(2) }{4\sqrt{2}+4+4 },\frac{4\sqrt{2}+4(4)+4(0) }{4\sqrt{2}+4+4 } )$

$(x,y)=(\frac{8+2\sqrt{2} }{2+\sqrt{2} }, \frac{4+4\sqrt{2} }{2+\sqrt{2} } )$

$(x,y)=(6-2\sqrt{2}, 4\sqrt{2} )$

Hence, coordinates of the incentre of the given triangle is $(6-2\sqrt{2}, 4\sqrt{2} )$.