In the triangle $ABC$ with vertices $A$$(2,3), B(4,-1)$ and $C(1,2),$ find the equation and length of altitude from the vertex $A$.
In the triangle $ABC$ with vertices $A$$(2,3), B(4,-1)$ and $C(1,2),$ find the equation and length of altitude from the vertex $A$.
Let $AD$ be the altitude of triangle $ABC$ from vertex $A$.
Accordingly, $AD \bot BC$
The equation of the line passing through point $(2,3)$ and having a slope of $1$ is $(y-3)=1(x-2)$
$\Rightarrow x-y+1=0$
$\Rightarrow y-x=1$
Therefore, equation of the altitude from vertex $A=y-x=1$
Length of $AD =$ Length of the perpendicular from $A (2, 3)$ to $BC$
The equation of $BC$ is
$(y+1)=\frac{2+1}{1-4}(x-4)$
$\Rightarrow(y+1)=-1(x-4)$
$\Rightarrow y+1=-x+4$
$\Rightarrow x+y-3=0$.....$(1)$
The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by
$d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$
On comparing equation $(1)$ to the general equation of line $A x+B y+C=0,$ we obtain $A=1$ $B =1,$ and $C =-3$
$\therefore$ Length of $AD =\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^{2}+1^{2}}}$ units $=\frac{|2|}{\sqrt{2}}$ units $=\frac{2}{\sqrt{2}}$ units $=\sqrt{2}$ units
Thus, the equation and length of the altitude from vertex $A$ are $y-x=1$ and $\sqrt{2}$ units respectively.
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