Coordinate Geometry
NCERT Exemplar Ch 07
CBSE_NCERT_EXEMPLAR_CH07
Grade 10
Question:
If the point $P(2, 1)$ lies on the line segment joining $A(4, 2)$ and $B(8, 4)$, then:
$AP = \dfrac{1}{3} AB$
$AP = PB$
$PB = \dfrac{1}{3} AB$
$AP = \dfrac{1}{2} AB$
Step-by-Step Solution
Key Concept: Check distances $AP, PB, AB$.
Stepwise Solution:
$AP = \sqrt{(4-2)^2 + (2-1)^2} = \sqrt{4+1} = \sqrt{5}$. [0.5 Mark]
$AB = \sqrt{(8-4)^2 + (4-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$. Hence $AP = \dfrac{1}{2} AB$. [0.5 Mark]
Marking Scheme:
• Calculating distances $AP = \sqrt{5}$ and $AB = 2\sqrt{5}$: 0.5 Mark
• Concluding $AP = \dfrac{1}{2} AB$: 0.5 Mark
Correct Answer: $AP = \dfrac{1}{2} AB$
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