Question

The equation to the sides of a triangle are $x-3y=0,4x+3y=5$ and $3x+y=0$.

Then line $3x-4y=0$ passes through

• A. incentre
• B. centroid
• C. circumference
• D. orthocentre of the triangle

Answer 1

The correct option is D

orthocentre of the triangle

Step 1: Calculate the slopes:

The given equation of lines are

$$\begin{gathered} L_1:x-3y=0 \\ {L}_2:4x+3y=5 \\ {L}_3:3x+y=0 \end{gathered}$$

Slope of the lines is:

$y=mx+c$

$$\begin{gathered} m_1=\frac{1}{3} \\ {m}_2=-\frac{4}{3} \\ {m}_3=-3 \end{gathered}$$

Here,

$\begin{array}{rcl}{m}_1{m}_3& =& \left(\frac{1}{3}\right)\left(-3\right)\\ & =& -1\end{array}$

So, $L_1\&L_3$ are perpendicular to each other.

Step 2: Calculate the required condition

Let $L:3x-4y=0$

Slope $m=\frac{3}{4}$.

Here, $m_{2}m=-1$.

So $L$is perpendicular to $L_2$.

The line $L$ passes through the origin and it is perpendicular to $L_2$.

So $L$ will pass through the orthocenter.

Hence, the line passes through the orthocenter and therefore option $\left(D\right)$ is the correct answer.