Question

Find the orthocenter of the triangle the equation of whose sides are $x+y=1,2x+3y=0$ and $4x-y+4=0$.

Answer 1

Let $AB,BC$ and $CA$ be $x+y=1,2x+3y=0$ and $4x-y+4=0$ respectively

Solving for $A$, we have $y=1-x$.

Hence $4x-y+4=0$

$4x+4=y\Rightarrow 4x+4=1-x$

$4x+x=1-4$

$5x=-3\Rightarrow x=-3/5$

$y=1-x=1+\frac{3}{5}=\frac{8}{5}$ $A(\frac{-3}{5},\frac{8}{5})$

Solving for $B,2x+3(1-x)=0$

$2x+3-3x=0$

$x=3$

$y=1-x=1-3=-2$ $B(3,-2)$

Altitude from $A$ passes through $A$ and perpendicular to $BC$.

Equation is $3x-2y+k=0$

$3\left(\frac{-3}{5}\right)-2\left(\frac{8}{5}\right)+k=0$ gives $k=5$.

Similarly from $B$, the altitude's equation will be

$x+4y+l=0$, passing through $(3,-2)$.

$3-8+l=0$ gives $l=5$.

The point of intersection of $3x-2y+5=0$ and $x+4y+5=0$ on solving is $(\frac{-15}{7},\frac{-5}{7})$

which is the orthocentre of the triangle.

$(\frac{-15}{7},\frac{-5}{7})$