Let AD be the altitude of triangle ABC from vertex A.
So, AD is perpendicular to BC
Given:
Vertices A (2, 3), B (4, –1) and C (1, 2)
Let slope of line BC = m1
m1 = (- 1 – 2)/(4 – 1)
m1 = -1
Let slope of line AD be m2
AD is perpendicular to BC
m1 × m2 = -1
-1 × m2 = -1
m2 = 1
The equation of the line passing through point (2, 3) andhaving a slope of 1 is
y – 3 = 1 × (x – 2)
y – 3 = x – 2
y – x = 1
Equation of the altitude from vertex A = y – x = 1
Length of AD = Length of the perpendicular from A (2, 3) toBC
Equation of BC is
y + 1 = -1 × (x – 4)
y + 1 = -x + 4
x + y – 3 = 0 …………………(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from apoint (x1, y1) is given by
Now compare equation (1) to the general equation of linei.e., Ax + By + C = 0, we get
Lengthof AD =
[where, A = 1, B = 1 and C = -3]
∴ The equation and the length of the altitude from vertex Aare y – x = 1 and
√2 units respectively.