Question

If $(0,1),(1,1)$ and $(1,0)$ are mid-points of sides of a triangle, find its incentre.

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

Answer 1

The correct option is D $(2-\sqrt{2},2-\sqrt{2})$

Given midpoints of the triangle are $(0,1),(1,1),(1,0)$.

From the figure we can say that $x_a=x_b=0,x_c=2$ and $y_a=y_c=0,y_b=2$.

Let the length of the opposite sides having coordinates $(x_a,y_a),(x_b,y_b),(x_c,y_c)$ be $a,b,c$ respectively.

Then $a=\sqrt{(x_b-x_c{)}^2+(y_b-y_c{)}^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$,

$b=\sqrt{(x_a-x_c{)}^2+(y_a-y_c{)}^2}=\sqrt{2^2+0^2}=\sqrt{4+0}=\sqrt{4}=2$,

$c=\sqrt{(x_b-x_a{)}^2+(y_b-y_a{)}^2}=\sqrt{0^2+2^2}=\sqrt{0+4}=\sqrt{4}=2$.

That is, $a=2\sqrt{2},b=2,c=2$

We know that the coordinates of the incentre is given by $(\frac{a{x}_a+b{x}_b+c{x}_c}{a+b+c},\frac{a{y}_a+b{y}_b+c{y}_c}{a+b+c})$

$=(\frac{2\cdot 2}{2\sqrt{2}+2+2},\frac{2\cdot 2}{2\sqrt{2}+2+2})$

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

$=(\frac{2}{2+\sqrt{2}},\frac{2}{2+\sqrt{2}})$

$=(\frac{2(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})},\frac{2(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})})$

$=(\frac{4-2\sqrt{2}}{4-2},\frac{4-2\sqrt{2}}{4-2})$

$=(\frac{4-2\sqrt{2}}{2},\frac{4-2\sqrt{2}}{2})$

$=(2-\sqrt{2},2-\sqrt{2})$

Thus the coordinates of the incentre are $(2-\sqrt{2},2-\sqrt{2})$