Two vertices of a triangle are (3, – 2) and (–2, 3) and its orthocenter is (–6, 1). Then the third vertex of this triangle can NOT lie on the line

Let $m_1=slope of$

$BH=\frac{1-(-2)}{-6-3}=\frac{-1}{3}$

$\Rightarrow$Slope of $AC=3$ 

$\therefore$Equation of AC is 

 $y-3=3(x+2)$ or $3x-y+9=0$       (1)

Let $m_2 =$slope of $CH=\frac{1-3}{-6+2}=\frac{1}{2}$

$\Rightarrow$Slope of AB  $=-2$

$\therefore$Equation of AB is $y+2=-2(x-3)$

or $2x+y-4=0$                                (2)

As A lies on (1) and (2), coordinates of A are $(– 1, 6)$and it does not lie on $5x+y=2$