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.