Question

Find the equations to the altitudes of the triangle whose angular points are A (2, -2), B (1, 1) and C (-1, 0).

Answer 1

$\text{AD, BE and CF are the three altitudes of the triangle}\text{We know,}SlopeofAD\times SlopeofBC=-1,\text{AD passes through A(2, -2)}SlopeofBE\times SlopeofAC=-1,\text{AD passes through B(1, 1)}SlopeofCF\times SlopeofAB=-1,\text{AD passes through C(-1, 0)}\text{Slope of Bc}=\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\Rightarrow \text{Slope of AD}=-2\text{Slope of AC}=\frac{0-(-2)}{-1-2}=\frac{2}{-3}=\frac{-2}{3}\Rightarrow \text{Slope of BE}=\frac{3}{2}\text{Slope of AB}=\frac{1+2}{1-2}=\frac{2}{-1}=-3\Rightarrow \text{Slope of CF}=\frac{1}{3}\text{So, for AD, we have}y-y_1=m(x-x_1)\Rightarrow y-(-2)=-2(x-2)\Rightarrow y+2=-2x+4\Rightarrow 2x+y-2=0\text{And, for BE, we have}y-y_1=m(x-x_1)\Rightarrow y-1=\frac{3}{2}(x-1)\Rightarrow [2y-3x+1=0]\Rightarrow 3x-2y-1=0\text{And, for CF, we have}y-y_1=m(x-x_1)\Rightarrow y-0=\frac{1}{3}(x+1)\Rightarrow x-3y+1=0$