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

Answer 1

Answer: (C)

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

$12x^2-20xy+7y^2=0$
Substitute $\frac{y}{x}=t,$
$7t^2-20t+12=0$
$t=\frac{20\pm \sqrt{400-336}}{14}$
$t=\frac{20\pm 8}{14}$
$t=2,\frac{6}{7}$
$\Rightarrow y=2x$ ---(1) $y=\frac{6}{7}x$ ---(2)
and $3y=2x+4$ ---(3)
from 1 & 3, $x=1$ & $y=2$
from 2& 3, $x=7$ & $y=6$
from 1 & 2, $x=0$ & $y=0$
centroid of $\Delta ABC=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
$=(\frac{8}{3},\frac{8}{3})$