WAEC Mathematics 2018 Paper

Concept:Use the two-point form of a straight line to determine its equation, then compare with the given equation to identify $p$.Explanation:Let $(x_1, y_1) = (4, 2)$ and $(x_2, y_2) = (-8, -2)$.The two-point form is:$$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$$Substitute the given points:$$\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}$$Simplify the denominators:$$\frac{y - 2}{-4} = \frac{x - 4}{-12}$$Cross-multiply:$$-12(y - 2) = -4(x - 4)$$Divide both sides by $-4$:$$3(y - 2) = x - 4$$Expand:$$3y - 6 = x - 4$$Add $6$ to both sides:$$3y = x + 2$$Compare with the given equation $3y = px + q$.The coefficient of $x$ is $1$, so $p = 1$.Answer:$p = 1$, which is option A.