The centroid of the triangle whose three sides are given by the combined equation $\left( x^{2} + 7xy + 2y^{2} \right)(y - 1) = 0$, is

Question

Dr. Abubacker · Accepted Answer

(c)
The sides of the triangle are $y = 1$ and the pair of lines $x^{2} + 7\ xy + {2y}^{2} = 0$
Clearly, one vertex is (0, 0) and the $y$-coordinates of each of the other two vertices is 1.
On putting $y = 1$ in the second equation, we get
$x^{2} + \ 7x + 2 = 0$
If $x_{1}$ and $x_{2}$ are the roots of this equation, then
$x_{1} + x_{2} = - 7$
$	herefore Centroid,\ G = \left( \frac{0 + x_{1} + x_{2}}{3},\frac{0 + 1 + 1}{3} ight)$
$= \left( - \frac{7}{3},\frac{2}{3} ight)$

Q. The centroid of the triangle whose three sides are given by the combined equation $\left( x^{2} + 7xy + 2y^{2} \right)(y - 1) = 0$, is

Explanation

Reviewed by Dr. Abubacker

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