Question

The vertices of a triangle are $A(0,b),B(0,0)$ and $C(a,0)$. If the medians through the vertices $A$ and $B$ are mutually perpendicular, then ${\left(\frac{a}{b}\right)}^2$ is equal to _____

Answer 1

Here,

Let AE and BF be medians of triangle ABC.

$\therefore$ E is the midpoint of BC.

$\therefore$ Coordinates of $E=(\frac{a+0}{2},\frac{0+0}{2})$

$=(\frac{a}{2},0)$

Similarly F is the midpoint of AC.

$\therefore$ Coordinates of $F=(\frac{a+0}{2},\frac{D+b}{2})$

$=(\frac{a}{2},\frac{b}{2})$

Now, slope of $AE,m_1=\frac{0-b}{a/2-0}=\frac{-b}{a/2}=\frac{-2b}{a}$

Slope of $BE,m_2=\frac{b/2-0}{a/2-0}=\frac{b/2}{a/2}=\frac{b}{a}$

Since AE and BF are mutually perpendicular

$\therefore m_1\cdot m_2=-1$

$\Rightarrow \frac{-2b}{a}\cdot \frac{b}{a}=-1$

$\Rightarrow -\frac{2b^2}{a^2}=-1$

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

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

$\Rightarrow {\left(\frac{a}{b}\right)}^2=2$.