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 = m_{1}

m_{1} = (- 1 – 2)/(4 – 1)

m_{1} = -1

Let slope of line AD be m_{2}

AD is perpendicular to BC

m_{1} × m_{2} = -1

-1 × m_{2} = -1

m_{2} = 1

The equation of the line passing through point (2, 3) and having 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) to BC

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 a point (x_{1}, y_{1}) is given by

Now compare equation (1) to the general equation of line i.e., Ax + By + C = 0, we get

Length of AD =

\( \dfrac{|1\times 2+1\times 3-3|}{\sqrt{1^2+1^2}}\)

= \( \dfrac{|2|}{\sqrt{2}}\) = \( \sqrt{2}\)

[where, A = 1, B = 1 and C = -3]

The equation and the length of the altitude from vertex A are y – x = 1 and \( \sqrt{2}\) units respectively.

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