The graph below shows the triangle ABC
We use points C(-2,6) and B(3,4) to find the slope of CB
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-6}{3-(-2)}=\frac{-2}{3+2}=-\frac{2}{5}[/tex]Now, we find the perpendicular slope to CB.
[tex]\begin{gathered} m\cdot m_1=-1 \\ m\cdot(-\frac{2}{5})=-1 \\ m=\frac{5}{2} \end{gathered}[/tex]The altitude has to pass through point A(1, -2). Let's use the point-slope formula to find the equation
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-(-2)=\frac{5}{2}(x-1) \\ y+2=\frac{5}{2}x-\frac{5}{2} \\ y=\frac{5}{2}x-\frac{5}{2}-2 \\ y=\frac{5}{2}x+\frac{-5-4}{2} \\ y=\frac{5}{2}x-\frac{9}{2} \end{gathered}[/tex]This altitude is perpendicular to the side CB.
