# Q : 17      In the triangle $ABC$ with vertices  $A(2,3)$, $B(4,-1)$ and  $C(1,2)$ , find the equation and length of altitude from the vertex $A$.

G Gautam harsolia

Let suppose foot of perpendicular is $(x_1,y_1)$
We can say that line passing through point $(x_1,y_1) \ and \ A(2,3)$  is perpendicular to line passing through point $B(4,-1) \ and \ C(1,2)$
Now,
Slope of line passing through point $B(4,-1) \ and \ C(1,2)$ is , $m' = \frac{2+1}{1-4}= \frac{3}{-3}=-1$
And
Slope of line  passing through point $(x_1,y_1) \ and \ (2,3)$ is , $m$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'}= 1$
Now,  equation of line passing through point  (2 ,3)  and slope with 1
$(y-3)=1(x-2)$
$x-y+1=0$                     -(i)
Now, equation line passing through point $B(4,-1) \ and \ C(1,2)$ is
$(y-2)=-1(x-1)$
$x+y-3=0$
Now, perpendicular distance of (2,3) from the $x+y-3=0$ is
$d= \left | \frac{1\times2+1\times3-3}{\sqrt{1^2+1^2}} \right |= \left | \frac{2+3-3}{\sqrt{1+1}} \right |= \frac{2}{\sqrt{2}}=\sqrt2$                -(ii)

Therefore, equation and length of the line is  $x-y+1=0$  and  $\sqrt2$   respectively

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