The vertices of $\triangle A B C$ are $A(2,3), B(4,-1)$ and $C(1,2)$ and $A M$ is the altitude.
Slope of $B C=\frac{2+1}{1-4}=\frac{3}{-3}=-1$
$\therefore \quad$ Slope of altitude $A M=1$
Now, altitude passes through $A(2,3)$ and has the slope 1 .


$\therefore \quad$ Equation of $A M$ is $y-3=(x-2)$
$x-y+3-2=0$
or $y-x=1$
Equation of $B C$ passing through $B(4,-1)$ and $C(1,2)$
$y+1=\frac{2-4}{1+1}(x-4) \Rightarrow y+1=\frac{-2}{2}(x-4)$
$y+1=-(x-4)$
or $x+y=3$ or $x+y-3=0$
$\therefore \quad$ Length of altitude $=A M$
$=$ Perpendicular distance from $A(2,3)$ to $B C$.
$=\frac{2+3-3}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\therefore \quad$ length of altitude $=\sqrt{2}$
Equation of altitude is $y-x=1$ and length of altitude $=\sqrt{2}$