Find the relationship between $a$ and $b$. 
If the medians $PT$ and $RS$ of a triangle with vertices $P(0,b),Q(0,0)\ \text{and}\ R(a,0)$ are perpendicular to each other,which of the following satisfies the relationship between $a$ and $b$?

$a.)\ 4b^2=a^2\\
\color{green}{b.)\ 2b^2=a^2}\\
c.)\ a=-2b\\
d.)\ a^2+b^2=0$
I found $S\left(0,\dfrac{b}{2}\right)$ and $T\left(\dfrac{a}{2},0\right)$ 
and as $PT$ perpendicular to $RS$.
$\dfrac{\frac{b}{2}}{-a}\times \dfrac{b}{\frac{-a}{2}}=-1\\~\\
b^2+a^2=0$
But the book is giving option $b.)$
I have studied maths up to $12th$ grade.
 A: Obviously, $\Delta PQR$ is a right triangle having vertices $P(0, b)$, $Q(0, 0)$ & $S(a, 0)$ The coordinates of points S & T can be easily determined as $S\left(0, \frac{b}{2}\right)$ & $T\left(\frac{a}{2}, 0\right)$.
Given that medians $PT$ & $RS$ are normal to each other, hence we have the following condition
$$\color{#0ae}{(\text{slope of median PT})\times (\text{slope of median RS})=-1}$$ 
Now setting the corresponding values, we get
$$\left(\frac{b-0}{0-\frac{a}{2}}\right)\times \left(\frac{\frac{b}{2}-0}{0-a}\right)=-1 \implies \frac{b^2}{a^2}=-1 \implies \color{#0b4}{a^2+b^2=0}$$ Since, $a$ & $b$ are real numbers, hence we have $a=0$ & $b=0$ satisfying the equality obtained above. Obviously, the result obtained above is same as you did. You are right. Certainly, there is some printing mistake in the option provided in your book. While the correct option is (d).        
The actual reason is that in a right triangle, two medians drawn from the acute angled vertices can not be normal to each other. 
A: $$\overrightarrow{PT}=\left(\frac a2, -b\right)$$
$$\overrightarrow{RS}=\left(-a,\frac b2\right)$$
therefore
$$\overrightarrow{PT}\cdot\overrightarrow{RS}=-\frac{a^2+b^2}2$$
Thus, you are right. Note that this is not a triangle. That is, the medians of a right triangle can not be perpendicular.
A: PQR is a right angled triangle. If hypotenuses are perpendicular then a = b =0, you are right. A trivial case.
