Question

Find the co-ordinates of the orthocenter of the triangle whose sides are along the lines $x+y-6=0,2x+y-4=0$ and $x+2y-5=0$

Answer 1

Let $PQ\to x+y=6$...(1)

$QR\to 2x+y=4$...(2)

$RP\to x+2y=5$...(3)

Equation of any line through Q can be put as linear combination of (1) and (2)

$A(x+y-6)+B(2x+y-4)=0$

$(A+2B)x+(A+B)y-6A-4B=0$…(4)

If H (ortho center ) has to lie on this line it must be perpendicular to (3)

Product of slopes is −1 , it translates to

$1\times (A+2B)+2\times (A+B)=0\Rightarrow 3A+4B=0$

Choose B=−3 and A will be 4.

Putting these values in (4)equation of QH becomes

$-2x+y-12=0$…(5)

On similar lines equation of RH can be found out as

$x-y+1=0$(6)

solving (5) and (6)

$x=-11$,$y=-10$

ortho center is $(-11,-10)$