Coordinate Geometry
NCERT Exemplar Ch 07
CBSE_NCERT_EXEMPLAR_CH07
Grade 10
Question:
If $P\left(\dfrac{a}{3}, 4\right)$ is the midpoint of the line segment joining the points $Q(-6, 5)$ and $R(-2, 3)$, then the value of $a$ is:
$-4$
$-12$
$12$
$-6$
Step-by-Step Solution
Key Concept: x-coordinate of midpoint $\dfrac{a}{3} = \dfrac{x_1 + x_2}{2}$.
Stepwise Solution:
$\dfrac{a}{3} = \dfrac{-6 + (-2)}{2} = \dfrac{-8}{2} = -4$. [0.5 Mark]
$a = 3 \times (-4) = -12$. [0.5 Mark]
Marking Scheme:
• Midpoint x-coordinate equation setup: 0.5 Mark
• Solving for $a = -12$: 0.5 Mark
Correct Answer: $-12$
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