Question

Find the orthocenter of the triangle formed by the lines
$x+2y=0,4x+3y-5=0,3x+y=0$

Answer 1

Given equations are:

$x+2y=0$......(1)

$4x+3y-5=0$.......(2)

$3x+y=0$..........(3)

Solving (1) and (2), vertex $A=(0,0)$

Solving (1) and (3), vertex $B=(2,-1)$

Equation of BC is $4x+3y-5=0$

AB is perpendicular to BC and passes through $A=(0,0)$

Equation of AB is $3x-4y=0$........(4)

BE is perpendicular to AC

Therefore equation of BE is $x-3y=k$

BE passes through $B=(2,-1)$

$2+3=k\Rightarrow k=5$

Equation of BE is $x-3y=5$........(5)

Solving (4) and (5)

Orthocenter is $O(-4,-3)$