Question

$\Delta ABC$ lies in the plane with $A=(0,0)$, $B=(0,1)$ and $C=(1,0)$. Points $M$ and $N$ are chosen on $AB$ and $AC$, respectively, such that $MN$ is parallel to $BC$ and $MN$ divides the area of $\Delta ABC$ in half. Find the coordinates of $M$.

Answer 1

We have,

Given points are

$A(0,0),B(0,1),C(1,0)$

Let the coordinates of $M(0,a),N(a,0)$

So,

According to given figure,

$MN\parallel BC$ So,

$SlopeoflineMN=SlopeoflineBC$

$\frac{0-a}{a-0}=\frac{0-1}{1-0}$

$\Rightarrow \frac{-a}{a}=\frac{-1}{1}$

$\Rightarrow a=a$

Now,

According to given question,

$ar\left(\Delta AMN\right)=\frac{1}{2}ar\left(\Delta ABC\right)$

$\frac{1}{2}\times a\times a=\frac{1}{2}\times \frac{1}{2}\times 1\times 1$

$a^2=\frac{1}{2}$

$a=\pm \frac{1}{\sqrt{2}}$

Then, Points are

$M(0,a)=M(\pm \frac{1}{\sqrt{2}},0)$

$N(a,0)=N(0,\pm \frac{1}{\sqrt{2}})$

Hence, this is the answer.