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Deepak Bora
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Find the value of p for which the points ( – 5, 1), (1, p) and (4, – 2) are collinear.

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An Important Question of M.L Aggarwal book of class 10 Based on Section Formula Chapter for ICSE BOARD.
In this question you have to find the value of p for which the given points are collinear.
This is the Question Number 30, Exercise 11 of M.L Aggarwal.

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1 Answer

  1. Let A(-5,1) divides the line joining (1,p) and (4,-2) in the ratio m:n

    Then by section formula, x = (mx2+nx1)/(m+n)

    -5 = (m×4+n×1)/(m+n)

    -5 = (4m+n)/(m+n)

    -5m-5n = 4m+n

    -9m = 6n

    m/n = -9/6 = -2/3 …(i)

    By section formula, y = (my2+ny1)/(m+n)

    1 = (m×-2+n×p)/(m+n)

    1 = (-2m+pn)/(m+n)

    m+n = -2m+pn

    3m = (p-1)n

    m/n = (p-1)/3 ….(ii)

    Equating (i) and (ii)

    (p-1)/3 = -2/3

    p-1 = -2

    p = -2+1 = -1

    Hence the value of p is -1.

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