waecmaths question:
The equation of the line through the (4, 2) and (–8, –2) is 3y = px + q, where p and q are constants. Find the value of p
Option A:
1
Option B:
2
Option C:
3
Option D:
9
waecmaths solution:
$\begin{align} & ({{x}_{1}},{{y}_{1}})=(4,2),\text{ }({{x}_{2}},{{y}_{2}})=(-8,-2) \\ & \text{Using two points form equation} \\ & \frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ & \frac{y-2}{x-4}=\frac{-2-2}{-8-4} \\ & \frac{y-2}{x-4}=\frac{-4}{-12} \\ & \frac{y-2}{x-4}=\frac{1}{3} \\ & 3(y-2)=x-4 \\ & 3y-6=x-4 \\ & 3y=x+2 \\ & \text{Comparing }3y=x+2\text{ with }3y=px+q \\ & p=1s \\\end{align}$
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