Co Ordinate Geometry - Quantitative Aptitude Problems and Solutions

    Condition of collinearity of three ponits
    Three points A(x1, y1), B(x2, y2 and C(x3, y3) are collinear if and only if are ΔABC = 0.
    A,B,C are colinear <=> x1(y2-y3) + x2(y3-y1) + x3(y1 - y2) = 0


  1. 21.   Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear.

    A.) 5 B.) 4
    C.) 7 D.) 1
    Answer: Option 'A'

    x1 = -2, x2 = 3, x3 = 6 and y1 = 5, y2 = k, y3 = -1

    Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0

          <=> -2 (k+1) + 3(4) + 6(5-k) = 0

          <=> -2k-2 + 12+30 -6k = 0

          <=> 40 - 8k = 0

          <=> -8k = -40

          <=> k = 5.



  2. 22.   Find the value of k for which the points A(-2, 6), B(5, k) and C(8, 3) are collinear.

    A.) 5 B.) 4
    C.) 7 D.) 1
    Answer: Option 'A'

    x1 = -2, x2 = 5, x3 = 8 and y1 = 6, y2 = k, y3 = 3

    Now Δ = 0 <=> -2(k - 3) + 5(3 - 6) + 8(6 - k) = 0

          <=> -2 (k-3) + 5(-3) + 8(6-k) = 0

          <=> -2k + 6 - 15 + 48 - 8k = 0

          <=> 39 - 10k = 0

          <=> k = -(39/10)

          <=> k = 3.9.



  3. 23.  If the A(2, 3), B(5, k), and C(6, 7) are collinear, then k = ?

    A.) 11 B.) 12
    C.) 18 D.) 6
    Answer: Option 'D'

    x1 = 2, x2 = 5, x3 = 6 and y1 = 3, y2 = k, y3 = 7

    Now Δ = 0 <=> 2(k - 7) + 5(7 - 3) + 6(3 - k) = 0

          <=> 1/2 [2 (k-7) + 5(4) + 6(3-k)] = 0

          <=> 2k - 14 + 20 + 18 - 6k = 0

          <=> 24 - 4k = 0

          <=> 4k = 24

          <=> k = 6.



  4. 24.  If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k = ?

    A.) 3 B.) -3
    C.) 4 D.) -4
    Answer: Option 'B'

    x1 = 1, x2 = 2, x3 = k and y1 = 2, y2 = 4, y3 = 6

    = 1(4 - 6) + 2 (6 - 2) + k(2 - 4)

    = -2 + 12 - 4 + 2k - 4k = 0

    = 6 - 2k = 0

    = -2k = 6

    = k = -3.




Quantitative Aptitude Topics



Quantitative Aptitude Topics


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