The centroid of the triangle formed by the pair of straight lines $12 x^2-20 x y+7 y^2=0$ and the line $2…
The centroid of the triangle formed by the pair of straight lines $12 x^2-20 x y+7 y^2=0$ and the line $2 x-3 y+4=0$ is :
$\left(-\frac{7}{3}, \frac{7}{3}\right)$
$\left(-\frac{8}{3}, \frac{8}{3}\right)$
$\left(\frac{8}{3}, \frac{8}{3}\right)$
$\left(\frac{4}{3}, \frac{4}{3}\right)$
Solution
The separate equation of pair of straight lines of $12 x^2-20 x y+7 y^2=0$ are $(6 x-7 y)=0$ and $(2 x-y)=0$
These the equation of sides are
$6 x-7 y=0$ $\ldots$ (i)
$2 x-y=0$ $\ldots$ (ii)
and $2 x-3 y+4=0$ $\ldots$ (iii)
On solving these equations, taking two at a time we get the vertices of a triangle which are $A(0,0) ; B(1,2)$ and $(7,6)$
$\therefore$ Centroid of a triangle is
$\left(\frac{0+1+7}{3}, \frac{0+2+6}{3}\right)$
i.e., $\quad\left(\frac{8}{3}, \frac{8}{3}\right)$