WAEC 2009 · Paper 2 · Q7

The coordinates of the vertices of triangle $ABC$ are $A(-2, 1)$ , $B(4, -2)$ and $C(1, 8)$ . If $D(x, y)$ is the foot of the perpendicular from $A$ to $BC$ , find:

(a)
an equation connecting $x$ and $y$ ;
(b)
the unit vector in the direction of $\overrightarrow{BC}$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Gradient of $BC$ : $\dfrac{8 - (-2)}{1 - 4} = -\dfrac{10}{3}$ .
Gradient of $AD$ , from $A(-2, 1)$ to $D(x, y)$ : $\dfrac{y - 1}{x + 2}$ .
$AD$ is perpendicular to $BC$ , so the product of the gradients is $-1$ : $\dfrac{y - 1}{x + 2} \times \left(-\dfrac{10}{3}\right) = -1$ .
Multiply out: $10(y - 1) = 3(x + 2)$ , so $10y - 10 = 3x + 6$ .
The equation is $10y - 3x - 16 = 0$ .

(b)

$\overrightarrow{BC} = (1 - 4)\mathbf i + (8 - (-2))\mathbf j$
$= -3\mathbf i + 10\mathbf j$ .
Its length is $\sqrt{(-3)^2 + 10^2} = \sqrt{109}$ .
Divide by the length: the unit vector is $\dfrac{1}{\sqrt{109}}(-3\mathbf i + 10\mathbf j)$ , about $-0.287\mathbf i + 0.958\mathbf j$ .

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