CAT Exam Model Paper 3 with solutions for free online practice

(c) Let A ( x${}_1$ , y${}_1$ ) be the third vertex, Let AD, BE, CF be the perpendiculars from the vertices on the opposite side BC, CA, AB respectively. Then the orthocentre O(0, 0) is the point of intersection of AD, BE and CF.
Since AD i.e. AD is perpendicular to BC.
∴ Slope of OA × slope of BC = − 1
or $\frac{{\mathit{y}}_1-0}{{\mathit{x}}_1-0}\times \frac{3-\left(-1\right)}{-2-5}=-1$
$\Rightarrow {\mathit{y}}_1=\frac{7{\mathit{x}}_1}{4}$...(1)
Again, since OB is perpendicular to CA
∴ $\frac{-1-0}{2-0}\times \frac{{\mathit{y}}_1-3}{{\mathit{x}}_1+2}=-1$
$\Rightarrow 2{\mathit{x}}_1+10={\mathit{y}}_1-3$
$\Rightarrow {\mathit{x}}_1=-52$$\left(\because {\mathit{y}}_1=\frac{7{\mathit{x}}_1}{4}\right)$
From eq (i), ${\mathit{y}}_1=\frac{7{\mathit{x}}_1}{4}=\frac{7\times \left(-52\right)}{4}=-91$
Hence, the required coordinates of the third vertex A are ( x${}_1$ , y${}_1$ ) = $\left(-52,-91\right)$