Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right angles by the lines $y^2-8 x y-9 x^2=0$. Then, the centoid of the triangle is

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Aicharya · Accepted Answer

$y^2-8 x y-9 x^2=0$ $y^2-9 x y+x y-9 x^2=0$ $(y-9 x)(y+x)=0$ The two given lines are $y=9 x$ and $y=-x$ Slope of line $y=9 x$ is 9 and slope of line $y=-x$ is -1 . Let $A B$ and $B C$ be the line perpendicular to $y=9 x$ and $y=-x$ respectively. Slope of $A B=\frac{-1}{9}$ and slope of line $B C$ is 1. Equation of $A B$ is $y-\beta=\frac{-1}{9}(x-\alpha)$ $9 y-9 \beta=-x+\alpha$ $\Rightarrow \quad x+9 y-\alpha-9 \beta=0$ $M$ is Mid-point of the $A B$ and $y=9 x$ So, $x+9(9 x)+\alpha+9 \beta$ $82 x=\alpha+9 \beta$ $x=\frac{\alpha+9 \beta}{82}$ Coordinate of $M$ is $\left(\frac{\alpha+9 \beta}{82}, \frac{9 \alpha+81 \beta}{82}\right)$. $M$ is Mid-point of $A B$, Let coordinate of $A$ be $h, k$ $\frac{\alpha+9 \beta}{82}=\frac{\alpha+h}{2}$ and $\frac{9 \alpha+81 \beta}{82}=\frac{\beta+k}{2}$ $\Rightarrow \quad 2 \alpha+18 \beta=82 \alpha+82 h$ $\Rightarrow \quad 82 h=-80 \alpha+18 \beta$ $\Rightarrow \quad h=\frac{-80 \alpha+18 \beta}{82}$ and $18 \alpha+162 \beta=82 \beta+82 k$ $\Rightarrow \quad 82 k=18 \alpha-8.0 \beta$ $\Rightarrow \quad k=\frac{18 \alpha-80 \beta}{82}$ Equation of $B C$ is $y-\beta=1(x-\alpha)$ $x-y=\alpha-\beta$ $N$ is intersection point of $B C$ and $y=-x$ $\therefore \quad x+x=\alpha-\beta$ $\begin{array}{ll}\Rightarrow & 2 x=\alpha-\beta \\ \Rightarrow & x=\frac{\alpha-\beta}{2}\end{array}$ $\Rightarrow \quad y=\frac{\beta-\alpha}{2}$ Coordinate of $N$ is $\left(\frac{\alpha-\beta}{2}, \frac{\beta-\alpha}{2}\right)$. $N$ is Mid-point of $B C$. Let coordinate of $C$ be $(a, b)$ $\frac{\alpha-\beta}{2}=\frac{\alpha+a}{2}$ and $\frac{\beta-\alpha}{2}=\frac{\beta+b}{2}$ $\Rightarrow a=-\beta$ and $b=-\alpha$ $\therefore$ Coordinate of $C$ is $(-\beta,-\alpha)$. Centroid of $A B C$ is $\left(\frac{h+a+\alpha}{3}, \frac{k+b+\beta}{3}\right)$ $=\left(\frac{1}{2}\left[\frac{-80 \alpha+18 \beta}{52}-\beta+\alpha\right], \frac{1}{2}\left[\frac{18 \alpha-80 \beta}{82}-\alpha+\beta\right]\right)$

$=\left(\frac{1}{3}\left[\frac{2 \alpha-64 \beta}{82}\right], \frac{1}{3}\left[\frac{-64 \alpha+2 \beta}{82}\right]\right)$ $=\frac{1}{123}(\alpha-32 \beta,-32 \alpha+\beta)$

Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right…

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