If two sides of a triangle are given by $3 x^2-5 x y+2 y^2=0$ and its orthocentre is $(2,1)$, then the equation of the third side of the triangle is

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Aicharya · Accepted Answer

Given pair equation of two sides of triangle, $ \begin{aligned} 3 x^2-5 x y+2 y^2 & =0 \\ \Rightarrow \quad(3 x-2 y)(x-y) & =0 \end{aligned} $ So, equation of sides are

Perpendicular line to the Eq. (i) $2 x+3 y+k=0$ which pass through the point $(2,1)$. $ \Rightarrow \quad 4+3+k=0 $ $\Rightarrow \quad k=-7$

Point of intersection of Eqs. (ii) and (iii) $ x=\frac{7}{5}, y=\frac{7}{5} \Rightarrow\left(\frac{7}{5}, \frac{7}{5}\right) $ Perpendicular line to the Eq. (2) is $ \begin{array}{ll} x+y+k=0 \text { which pass through }(2,1) \\ \Rightarrow & 2+1+k=0 \\ \Rightarrow & k=-3 \end{array} $

Point of intersection of Eqs. (iv) and (i) $ \begin{aligned} x+y-3 & =0 \\ 3 x-2 y & =0 \Rightarrow x=\frac{6}{5}, y=\frac{9}{5} \Rightarrow\left(\frac{6}{5}, \frac{9}{5}\right) \end{aligned} $ So, third side of triangle pass through these point $\left(\frac{7}{5}, \frac{7}{5}\right)$ and $\left(\frac{6}{5}, \frac{9}{5}\right)$ is $ \begin{array}{ll} \Rightarrow & y-\frac{7}{5}=\frac{\frac{9}{5}-\frac{7}{5}}{\frac{6}{5}-\frac{7}{5}}\left(x-\frac{7}{5}\right) \\ \Rightarrow & y-\frac{7}{5}=\frac{2}{-1}\left(x-\frac{7}{5}\right) \\ \Rightarrow & \frac{5 y-7}{5}=\frac{-10 x+14}{5} \\ \Rightarrow & 10 x+5 y=21 . \end{array} $

If two sides of a triangle are given by $3 x^2-5 x y+2 y^2=0$ and its orthocentre is $(2,1)$, then the…

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