Question

The medians $AD$ and $BE$ of a triangle with vertices $A=(0,b)$, $B=(0,0)$ and $C=(a,0)$ are perpendicular to each other, if

• A. $a=\sqrt{2}b$
• B. $a=-\sqrt{2}b$
• C. Both $\text{a}$ and $\text{b}$
• D. None of these

Answer 1

The correct option is C

Both $\text{a}$ and $\text{b}$

Explanation for the correct option:

Step 1: Find the coordinates of point $D$ and $E$.

It is given that $A=(0,b)$, $B=(0,0)$ and $C=(a,0)$.

Since, $AD$ is the median. So, $D$ is the midpoint of $BC$ and $BE$ is also a median. So, $E$ is the midpoint of $AC$.

So, the coordinate of $D$ can be given as follows:

$$\begin{gathered} D=\left(\frac{0+a}{2},\frac{0+0}{2}\right) \\ \Rightarrow D=\left(\frac{a}{2},0\right) \end{gathered}$$

Also, the coordinate of $E$ can be given as follows:

$$\begin{gathered} E=\left(\frac{a+0}{2},\frac{0+b}{2}\right) \\ \Rightarrow E=\left(\frac{a}{2},\frac{b}{2}\right) \end{gathered}$$

Therefore, the coordinates of point $D$ and $E$ are $\left(\frac{a}{2},0\right)$ and $\left(\frac{a}{2},\frac{b}{2}\right)$ respectively.

Step 2: Find the relation between $a$ and $b$.

Compute the slope $m_1$ of line $BE$.

$$\begin{gathered} m_1=\frac{{\displaystyle \frac{b}{2}}-0}{{\displaystyle \frac{a}{2}}-0} \\ \Rightarrow m_1=\frac{{\displaystyle b}}{{\displaystyle a}} \end{gathered}$$

Now, Compute the slope $m_2$ of line $AD$.

$$\begin{gathered} m_2=\frac{{\displaystyle 0}-b}{{\displaystyle \frac{a}{2}}-0} \\ \Rightarrow m_2=\frac{{\displaystyle -2b}}{{\displaystyle a}} \end{gathered}$$

Since, the lines $BE$ and $AD$ are perpendicular to each other.

We know that, the product of slopes of perpendicular lines is equal to $-1$.

Therefore,

$$\begin{gathered} m_1·m_2=-1 \\ \Rightarrow \frac{b}{a}·\frac{-2b}{a}=-1 \\ \Rightarrow \frac{2b^2}{a^2}=1 \\ \Rightarrow a^2=2b^2 \\ \Rightarrow a=\pm \sqrt{2}b \end{gathered}$$

Therefore, the relation between $a$ and $b$ are $a=\pm \sqrt{2}b$

Hence, option $C$ is the correct answer.