Straight Lines and Pair of Straight Lines

The equation $3 x^2-5 x y+2 y^2=0$ can be written as two factors: $(3 x-2 y)(x-y)=0$.This means the two sides of the triangle are $3 x-2 y=0$ and $x-y=0$.The slope of $3 x-2 y=0$ is $m_1=\;\frac{3}{2}$, and the slope of $x-y=0$ is $m_2=1$.To find where the triangle's heights (altitudes) go from the orthocentre at $(2,1)$, we need the slope of each altitude. The altitude to $3 x-2 y=0$ will have a slope that is the negative reciprocal of $\;\frac{3}{2}$, which is $-\;\frac{2}{3}$.The altitude to $x-y=0$ will have a slope that is the negative reciprocal of 1 , which is -1 .Write the equation for the altitude with slope $-\;\frac{2}{3}$ that passes through $(2,1)$ :$\;y-1=-\;\frac{2}{3}(x-2)$$\;\Rightarrow 3 y-3=-2 x+4$$\;\Rightarrow 2 x+3 y=7$Write the equation for the altitude with slope -1 that passes through $(2,1)$ :$\;y-1=-1(x-2)$$\;\Rightarrow y-1=-x+2$$\;\Rightarrow x+y-3=0$Find where each altitude meets the opposite side. First, where $2 x+3 y=7$ (altitude) meets$x-y=0 \;\text{ (side): }\;$Since $x=y$, substitute it into $2 x+3 x=7$, so $5 x=7$ which means $x=\;\frac{7}{5}$.The intersection is at $\left(\;\frac{7}{5}, \;\frac{7}{5}\right)$.From $3 x-2 y=0, x=\;\frac{2}{3} y$.Put this in $x+y-3=0$ to get $\;\frac{2}{3} y+y=3 \implies \;\frac{5}{3} y=3 \implies y=\;\frac{9}{5}$.When $y=\;\frac{9}{5}, x=\;\frac{6}{5}$, so the intersection point is $\left(\;\frac{6}{5}, \;\frac{9}{5}\right)$.The third side of the triangle must be the line connecting these two intersection points: $\left(\;\frac{7}{5}, \;\frac{7}{5}\right)$ and $\left(\;\frac{6}{5}, \;\frac{9}{5}\right)$.Find the slope of this line: $m=\;\frac{\frac{9}{5}-\frac{7}{5}}{\frac{6}{5}-\frac{7}{5}}=\frac{\frac{2}{5}}{\frac{-1}{5}}=-2$.The equation for this line, using point-slope form with the point $\left(\;\frac{7}{5}, \;\frac{7}{5}\right)$ and slope -2 , is:$\;y-\;\frac{7}{5}=-2 \left(x-\;\frac{7}{5}\right)$$\;\Rightarrow 5 y-7=-10 x+14$$\;\Rightarrow 10 x+5 y-21=0$So, the equation of the third side is $10 x+5 y-21=0$.