Centroid of a triangle, the equation of whose sides are $12 x^{2} - 20 x y + 7 y^{2} = 0$ and $2 x - 3 y + 4 = 0$ is

(\frac{8}{3}, \frac{8}{3})

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(\frac{4}{3}, \frac{4}{3})

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(2, 2)

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(1, 1)

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Solution

The correct option is B $(\frac{8}{3}, \frac{8}{3})$
$⟶$ Equation of sides $\Rightarrow$
$12 x^{2} - 20 x y + 7 y^{2} = 0$
$12 x^{2} - 6 x y - 14 x y + 7 y^{2} = 0$
$6 x (2 x - y) - 7 y (2 x - y) = 0$
$(6 x - 7 y) (2 x - y) = 0$
for $B \to$ we need to solve
$6 x - 7 y = 0$ & $2 x - 3 y + 4 = 0$
$y = \frac{6 x}{7}$
substituting $y$ in $2 x - 3 y + 4 = 0$
$2 x - 3 (\frac{6 x}{7}) + 4 = 0$
$2 x - \frac{18 x}{7} + 4 = 0 \Rightarrow 14 x - 18 x + 28 = 0$
$4 x = 28$
$x = 7$
$y = \frac{6 x}{7} = \frac{6}{7} 7 = 6$
$∴ B = (7, 6)$
for $C \to$ we need to solve
$2 x - 3 y + 4 = 0$ & $2 x - y = 0$
$y = 2 x$
substituting $y = 2 x$ in $2 x - 3 y + 4 = 0$
$2 x - 3 (2 x) + 4 = 0$
$2 x - 6 x + 4 = 0$
$4 x = 4$
$x = 1$
$y = 2 x$
$= 2 (1)$
$y = 2$
$∴ C \equiv (1, 2)$

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