The coordinates of the vertices of triangle ABC are A(-2, 1), B(4, -2) and C(1, 8) respectively. If D(x, y) is the foot perpendicular from A to BC, find
(a) an equation connecting x and y ;
(b) the unit vector in the direction of BC.
Gradient of BC = \(\frac{8 + 2}{1 - 4} = -\frac{10}{3}\)
Gradient of DC = \(\frac{8 - y}{1 - x}\)
BDC is a straight line , so
Gradient of BC = Gradient of DC.
\(\frac{10}{-3} = \frac{8 - y}{1 - x}\)
\(-3(8 - y) = 10(1 - x) \implies 3y - 24 = 10 - 10x\)
\(3y + 10x = 34\)
(b) \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}\)
= \((i + 8j) - (4i - 2j)\)
= \(-3i + 10j\)
\(|\overrightarrow{BC}| = \sqrt{(-3)^{2} + (10)^{2}}\)
= \(\sqrt{109}\)
Unit vector in the direction of BC = \(\frac{(-3i + 10j)}{\sqrt{109}}\)
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