The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
Using the two - point from
\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)
\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)
\(\frac{-12(y -2)}{-4}\) = x - 4
3(y -2) = x -4
3y - 6 = x - 4
3y = x - 4 + 6
3y = x + 2...
By comparing the equations;
3y = px + , p = 1
Using the two - point from
\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)
\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)
\(\frac{-12(y -2)}{-4}\) = x - 4
3(y -2) = x -4
3y - 6 = x - 4
3y = x - 4 + 6
3y = x + 2...
By comparing the equations;
3y = px + , p = 1