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Straight Lines and Pair of Straight Lines question

2019 · 11 Jan · Shift 2 · Q40
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Straight Lines and Pair of Straight Lines question

2019 · 11 Jan · Shift 2 · Q40

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is :
  1. A
    5x + 3y – 11 = 0
  2. B
    5x – 3y + 1 = 0
  3. C
    3x – 5y + 7 = 0
  4. D
    3x + 5y – 13 = 0
View written solutionFree

Correct answer: B

  1. Understand the figure

    In parallelogram ABDCABDCABDC, the vertices are given in the order A→B→D→CA \to B \to D \to CA→B→D→C.

    So the diagonal ADADAD is the line joining points AAA and DDD.

  2. Use the parallelogram property to find DDD

    In a parallelogram, diagonals bisect each other. Equivalently, A⃗+D⃗=B⃗+C⃗\vec{A} + \vec{D} = \vec{B} + \vec{C}A+D=B+C Hence, D=B+C−A.D = B + C - A.D=B+C−A.

    Given: A(1,2),B(3,4),C(2,5)A(1,2),\quad B(3,4),\quad C(2,5)A(1,2),B(3,4),C(2,5)

    Therefore, D=(3+2−1,  4+5−2)=(4,7).D = (3+2-1,\;4+5-2) = (4,7).D=(3+2−1,4+5−2)=(4,7).

  3. Find the equation of line ADADAD

    The line passes through A(1,2)A(1,2)A(1,2) and D(4,7)D(4,7)D(4,7).

    Slope of ADADAD: m=7−24−1=53.m = \frac{7-2}{4-1} = \frac{5}{3}.m=4−17−2​=35​.

    Using point-slope form through A(1,2)A(1,2)A(1,2): y−2=53(x−1).y-2 = \frac{5}{3}(x-1).y−2=35​(x−1).

    Multiply by 333: 3y−6=5x−53y - 6 = 5x - 53y−6=5x−5

    Rearranging, 5x−3y+1=0.5x - 3y + 1 = 0.5x−3y+1=0.

  4. Match with the options

    The required equation is: 5x−3y+1=0\boxed{5x - 3y + 1 = 0}5x−3y+1=0​

    This matches Option B.

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