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.

$\begin{array}{1 1}(A)\;\text{equation} = x-y=1 \: \text{and length is } \sqrt 2\: \text{units} \\(B)\; \text{equation} = x+y=1 \text{and length is } 2\: \text{units} \\(C)\; \text{equation} = -x+y+1=0 \: \text{and length is } 2\sqrt 2 \text{ units} \\(D)\;\text{equation} = x-y=-1 \text{ and length is } \large\frac{1}{\sqrt 2} \text{ units} \end{array}$

Toolbox:
• Equation of a line passing through $(x_1,y_1)$ and having slope $m$ is $y-y_1=m(x-x_1)$
• The perpendicular distance from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is given by $d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
Let AD be the altitude of the triangle from vertex A.
$\therefore AD \perp BC$
Hence equation of the line passing through the point (2,3) (4,-1) and having a slope 1 is
$(y-3)=1(x-2)$ $\Rightarrow x-y+1=0$
Length of AD = length of the perpendicular from A(2,3) to BC.
Slope of the line is $\large\frac{y_2-y_1}{x_2-x_1}$$= \large\frac{2+1}{1-4}$$=-1$
$\therefore$ The equation of BC is
$(y+1)=-1(x-4)$
$\Rightarrow (y+1)=-1(x-4)$
$\Rightarrow x+y-3=0$------------(1)
Now the perpendicular distance (d) of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is $d = \large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
On comparing the above equation with equation (1), we get
A=1, B=1 and C=-3
$\therefore$ length of the altitude $AD = \large\frac{|1 \times 2 +1 \times 3-3|}{\sqrt {1^2+1^2}}$
$= \large\frac{|2|}{\sqrt 2}$ units.
$\sqrt 2$ units.
Hence the equation of the altitude from vertex A is $x-y=1$ and length of the altitude = $\sqrt 2$ units.
edited May 24, 2014