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Two points $(4, p)$ and $(0, q)$ lie on a straight line having a slope of $3/4$. The value of $( p – q)$ is

1. $-3$
2. $0$
3. $3$
4. $4$

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For two points $(x_1,y_1)$ and $(x_2,y_2)$ on a line, Slope of line $= \dfrac{y_2-y_1}{x_2-x_1}$

so ,$\dfrac{q-p}{0-4} = \dfrac{3}{4}$

$\therefore (p-q)=3$

Correct Answer: $C$
by Veteran (57k points)
edited