Question

The mid points of the sides of a triangle ABC are (1/2,1/2), (1/2,0), (0,1/2). Then the incentre of triangle ABC is:

• A. $(\frac{1}{3},\frac{1}{3})$
• B. $(\frac{1}{2+\sqrt{2}}.\frac{1}{2+\sqrt{2}})$
• C. $(0,0)$
• D. $(1,2)$

Answer 1

Answer: (A)

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

We have,

$A(x_1,y_1),B(x_2,y_2)\text{and}(x_3,y_3)$ be the vertices of the triangle$ABC$.

And $P(\frac{1}{2},\frac{1}{2}),B(\frac{1}{2},0)andC(0,\frac{1}{2})$ be the midpoints of sides $AB,BCandCA$ respectively.

Since. $P$ is the mid point of side $AB$

Therefore,

$\frac{x_1+x_2}{2}=\frac{1}{2}\Rightarrow x_1+x_2=1......\left(1\right)$

$\frac{y_1+y_2}{2}=\frac{1}{2}\Rightarrow y_1+y_2=1......\left(2\right)$

Since,$Q$ is the midpoint of side $BC$

$\frac{x_2+x_3}{2}=\frac{1}{2}\Rightarrow x_2+x_3=1......\left(3\right)$

$\frac{y_2+y_3}{2}=0\Rightarrow y_2+y_3=0.......\left(4\right)$

Since, $R$ is the midpoint of side $CA$

$\frac{x_3+x_1}{2}=0\Rightarrow x_3+x_1=0......\left(5\right)$

$\frac{y_3+y_1}{2}=\frac{1}{2}\Rightarrow y_3+y_1=1......\left(6\right)$

On adding equation (1), (3) and (5) to and we get,

$x_1+x_2+x_2+x_3+x_3+x_1=1+1+0$

$2(x_1+x_2+x_3)=2$

$x_1+x_2+x_3=1$

On adding equation (2), (4) and (6) to and we get,

$y_1+y_2+y_2+y_3+y_3+y_1=1+0+1$

$2(y_1+y_2+y_3)=2$

$y_1+y_2+y_3=1$

Then, we know that,

In centre of triangle of

$(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$(x,y)=(\frac{1}{3},\frac{1}{3})$

Hence, this is the answer.