Question

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.

Answer 1

Let AD be the altitude of triangle ABC from vertex A.
Accordingly, $AD⊥BC$

The equation of the line passing through point (2, 3) and having a slope of 1 is
$(y \- 3) = 1(x \- 2)$
$⇒ x- y + 1 = 0$
$⇒ y \- x = 1$
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

$(y+1)=\frac{2+1}{1-4}(x-4)$
$⇒(y+1)=-1(x-4)$
$⇒y+1=-x+4$
$⇒x+y-3=0.......(1)$

The perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}$
On comparing equation (1) to the general equation of line $Ax + By + C = 0$, we obtain $A = 1, B = 1$, and $C = -3.$
∴ Length of $AD=\frac{\left|1\times2+1\times3-3\right|}{\sqrt{1^2+1^2}}$ units

$=\frac{\left|2\right|}{\sqrt2}$ units

$=\frac{2}{\sqrt2}$ units

$=\sqrt2$ units
Thus, the equation and the length of the altitude from vertex A are $y \- x = 1$ and $\sqrt2$ units respectively.