Question

The orthocentre of the triangle formed by the lines $x+y=1,2x+3y=6$ and $4x-y+4=0$ lies in

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. fourth quadrant

Answer 1

The correct option is A first quadrant

Let $H(h,k)$ be orthocentre.

$\Rightarrow ($slopes of $AH).($slope of $BC)$

$\Rightarrow \left(\frac{k-\frac{16}{7}}{h+\frac{3}{7}}\right).(-1)=-1$

$\Rightarrow k-\frac{16}{7}=h+\frac{3}{7}$

$\Rightarrow h-k=-\frac{19}{7}$ ...(i)

Also, $($slope of $CH).($slopes of $AB)=-1$

$\Rightarrow \frac{k-4}{h+3}.\left(4\right)=-1\Rightarrow 4k-16=-h-3$

$\Rightarrow h+4k=13$ ...(ii)