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.
Explanation
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}})