Question

The centroid of the triangle formed by the pair of straight lines $12x^2-20xy+7y^2=0$ and the line $2x-3y+4=0$ is

• A. $\left(\frac{-7}{3},\frac{7}{3}\right)$
• B. $\left(\frac{8}{3},\frac{8}{3}\right)$
• C. $\left(\frac{-8}{3},\frac{8}{3}\right)$
• D. $\left(\frac{4}{3},\frac{4}{3}\right)$

Answer 1

The correct option is B

$\left(\frac{8}{3},\frac{8}{3}\right)$

Explanation for the correct option:

Finding the centroid:

Given the equation of pair of lines is $12x^2-20xy+7y^2=0$.

$$\begin{gathered} \Rightarrow 12x^2-6xy-14xy+7y^2=0 \\ \Rightarrow 6x(2x-y)-7y(2x-y)=0 \\ \Rightarrow (2x-y)(6x-7y)=0 \\ \Rightarrow y=2x;6x=7y \end{gathered}$$

The three lines of triangle are $y=2x,6x=7y$ and $2x-3y+4=0$.

Solve equation $y=2x,6x=7y$.

$\Rightarrow \left(x_1,y_1\right)=\left(0,0\right)$

Solve the equation $y=2x$and $2x-3y+4=0$.

$\Rightarrow \left(x_2,y_2\right)=\left(1,2\right)$

Solve the equation $6x=7y$and $2x-3y+4=0$.

$\Rightarrow \left(x_2,y_2\right)=\left(7,6\right)$

So, the vertices of triangle are $(0,0),(1,2),(7,6)$.

The centroid of the triangle formed by vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by

$x=\frac{x_1+x_2+x_3}{3}$ and $y=\frac{y_1+y_2+y_3}{3}$

Substitute the values.

$\begin{array}{rcl}\therefore x& =& \frac{0+1+7}{3}\\ & =& \frac{8}{3}\end{array}$

and

$\begin{array}{rcl}y& =& \frac{0+2+6}{3}\\ & =& \frac{8}{3}\end{array}$

Hence, the centroid of the triangle is $\left(\frac{8}{3},\frac{8}{3}\right)$.

Therefore, the correct answer is option (B).