Question

The equations of the two sides of a triangle are $3x-2y+6=0$ and $4x+5y-20=0$ respectively. If the orthocentre of the triangle is (1, 1), find the equation of third side.

Answer 1

Let the equation of sides AB and AC of $\Delta ABC$ be $3x-2y+6=0$ and $4x+5y-20=0,$ respectively.

Let H(1,1) be the orthocentre of $\Delta ABC,$ where the altitudes AL,BM and CN intersect each other.

$\therefore$ BM passes through the point (1, 1) and perpendicular to AC.

$\therefore$ The equation of BM is $5x-4y=\lambda$

It passes through the point (1, 1), then $\lambda =1$

$\therefore$ Equation of BM is $5x-4y=1.$

Vertex B of $\Delta ABC$ is the intersection point of sides AB and BM.

On solving these equations, we get

$x=-13andy=\frac{-33}{2}$

$\therefore$ Coordinates of B are $(13,-\frac{33}{2}).$

Simliarly, CN is perpendicular to AB.

$\therefore$ The equation of CN is $2x+3y=\lambda$

It is passes through the point (1, 1), then $\lambda =5$

$\therefore$ Equation of CN is $2x+3y=5$

Vertex C of $\Delta ABC$ is the point of intersection of sides AC and CN.

On solving these equations of sides AC and CN, we get $x=\frac{35}{2}$ and $y=-10$

$\therefore$ Coordinates of B are $(\frac{35}{2},-10)$

Thus, the coordinates of B and C are $(-13,-\frac{33}{2})$ and $(\frac{35}{2}-10)$ respectively,

Hence, the equation of side BC is $y+\frac{33}{2}=\frac{-10+\frac{33}{2}}{\frac{35}{2}+13}(x+13)$ $[usingy-y_1=\left(\frac{y_2-y_2}{x_2-x_1}\right)(x-x_1\left)\right]$

$\Rightarrow \frac{2y+33}{2}=\frac{-20+33}{35+26}(x+13)\Rightarrow 2y+33=\frac{26}{61}(x+13)$

$\therefore 26x-122y-1675=0$