The vertices of $△ A B C$ are $A (2, 1), B (6, - 1)$ and $C (4, 11)$ . Find the equation of the straight line along the altitude from the vertex $A$ .

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Solution

Slope of $B C = \frac{11 + 1}{4 - 6} = - 6$
Since the line $A D$ is perpendicular to the line $B C$ , slope of $A D = \frac{1}{6}$
$∴$ Equation of $A D$ is $y - y_{1} = m (x - x_{1})$
$y - 1 = \frac{1}{6} (x - 2) \Rightarrow$ $6 y - 6 = x - 2$
$∴$ Equation of the required straight line is $x - 6 y + 4 = 0$

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