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Straight Lines and Pair of Straight Lines question

2021 · 17 Mar · Shift 1 · Q23
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Straight Lines and Pair of Straight Lines question

2021 · 17 Mar · Shift 1 · Q23

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
In a triangle PQR, the co-ordinates of the points P and Q are (−-− 2, 4) and (4, −-− 2) respectively. If the equation of the perpendicular bisector of PR is 2x −-− y + 2 = 0, then the centre of the circumcircle of the Δ\DeltaΔ PQR is :
  1. A
    (−-− 1, 0)
  2. B
    (1, 4)
  3. C
    (0, 2)
  4. D
    (−-− 2, −-− 2)
View written solutionFree

Correct answer: D

  1. Given points

    P(−2,4),Q(4,−2)P(-2,4), \quad Q(4,-2)P(−2,4),Q(4,−2)

    The centre of the circumcircle of triangle PQRPQRPQR is the intersection point of the perpendicular bisectors of its sides.

  2. Perpendicular bisector of PQPQPQ

    First find the midpoint of PQPQPQ: MPQ=(−2+42,4+(−2)2)=(1,1)M_{PQ} = \left(\frac{-2+4}{2},\frac{4+(-2)}{2}\right) = (1,1)MPQ​=(2−2+4​,24+(−2)​)=(1,1)

    Slope of PQPQPQ: mPQ=−2−44−(−2)=−66=−1m_{PQ} = \frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1mPQ​=4−(−2)−2−4​=6−6​=−1

    Therefore, slope of the perpendicular bisector of PQPQPQ is: m=1m = 1m=1

    Equation through (1,1)(1,1)(1,1): y−1=1(x−1)y-1 = 1(x-1)y−1=1(x−1) y=xy = xy=x

  3. Perpendicular bisector of PRPRPR

    It is given as: 2x−y+2=02x - y + 2 = 02x−y+2=0 y=2x+2y = 2x + 2y=2x+2

  4. Find circumcentre = intersection of the two perpendicular bisectors

    Solve: y=xy=xy=x and y=2x+2y=2x+2y=2x+2

    So, x=2x+2x = 2x+2x=2x+2 −x=2-x = 2−x=2 x=−2x = -2x=−2

    Hence, y=x=−2y = x = -2y=x=−2

    Therefore, the circumcentre is: (−2,−2)(-2,-2)(−2,−2)

  5. Check with options

    This matches Option D.

  6. Compare with stored correct answer

    Stored correct answer is D, which agrees with the derived result.

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