10th Class Mathematics Coordinate Geometry Question Bank MCQs - Coordinate Geometry

  • question_answer
    If the point \[P(2,1)\] lies on the line segment joining point \[A(4,2)\] and \[B(8,4),\] then: (ncert exemplar)

    A) \[AP=\frac{1}{3}AB\]

    B) \[AP=PB\]

    C) \[PB=\frac{1}{3}AB\]

    D) \[AP=\frac{1}{2}AB\]

    Correct Answer: D

    Solution :

    [d] Given that, the point \[P(2,1)\] lies on the line segment joining the points \[A(4,2)\] and \[B(8,4)\].
    Now, distance between \[A(4,2)\] and \[P(2,1)\].
    \[AP=\sqrt{{{(2-4)}^{2}}+{{(1-2)}^{2}}}=\sqrt{4+1}=\sqrt{5}\]
    Distance between \[A(4,2)\]and \[B(8,4)\]
    \[AB=\sqrt{{{(8-4)}^{2}}+{{(4-2)}^{2}}}\]
    \[=\sqrt{{{(4)}^{2}}+{{(2)}^{2}}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\]
    Distance between \[B(8,4)\] and \[P(2,1),\]
    \[BP=\sqrt{{{(8-2)}^{2}}+{{(4-1)}^{2}}}=\sqrt{{{6}^{2}}+{{3}^{2}}}\]
    \[=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\]
    \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,AB=2\sqrt{5}=2AP\,\,\Rightarrow \,\,AP=\frac{AB}{2}\]
    Hence, required condition is \[AP=\frac{AB}{2}\]


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