Question

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

• A. I quadrant
• B. II quadrant
• C. III quadrant
• D. IV quadrant

Answer 1

The correct option is A I quadrant

Coordinates of A and B are (-3, 4) and $(-\frac{3}{5},\frac{8}{5})$​ if orthocenter P(h, k)

Then, (slope of PA)$\times$ (slope of BC) = - 1
$\frac{k-4}{h+3}\times 4=-1$

$\Rightarrow$ 4k - 16 = -h - 3
$\Rightarrow$ h + 4k = 13....(i)

and slope of PB$\times$ slope of AC = - 1
$\Rightarrow \frac{k-\frac{8}{5}}{h+\frac{3}{5}}\times -\frac{2}{3}=-1$
$\Rightarrow \frac{5k-8}{5h+3}\times \frac{2}{3}=1$
$\Rightarrow$ 10k - 16 = 15th + 9
15th - 10k + 25 = 10
3h - 2k + 5 = 0 ...(ii)
Solving Eqs. (i) and (ii), we get $h=\frac{3}{7},k=\frac{22}{7}$
Hence, orthocentre lies in I quadrant.