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Straight Lines and Pair of Straight Lines question

2020 · 4 Sep · Shift 2 · Q30
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Straight Lines and Pair of Straight Lines question

2020 · 4 Sep · Shift 2 · Q30

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to –4, then a value of k is :
  1. A
    14\sqrt {14}14​
  2. B
    -4
  3. C
    –2
  4. D
    15\sqrt {15}15​
View written solutionFree

Correct answer: B

  1. Let the points be P(1,4),Q(k,3).P(1,4), \quad Q(k,3).P(1,4),Q(k,3).

  2. The midpoint of PQPQPQ is M(1+k2,4+32)=(1+k2,72).M\left(\frac{1+k}{2},\frac{4+3}{2}\right)=\left(\frac{1+k}{2},\frac{7}{2}\right).M(21+k​,24+3​)=(21+k​,27​).

  3. Slope of PQPQPQ is mPQ=3−4k−1=−1k−1.m_{PQ}=\frac{3-4}{k-1}=\frac{-1}{k-1}.mPQ​=k−13−4​=k−1−1​.

  4. Therefore, slope of the perpendicular bisector is the negative reciprocal: m=k−1.m = k-1.m=k−1.

  5. Equation of the perpendicular bisector through midpoint MMM: y−72=(k−1)(x−1+k2).y-\frac{7}{2}=(k-1)\left(x-\frac{1+k}{2}\right).y−27​=(k−1)(x−21+k​).

  6. Its yyy-intercept is obtained by putting x=0x=0x=0: y−72=(k−1)(−1+k2).y-\frac{7}{2}=(k-1)\left(-\frac{1+k}{2}\right).y−27​=(k−1)(−21+k​). So,

    =\frac{7-(k^2-1)}{2} =\frac{8-k^2}{2}.$$

  7. Given that the yyy-intercept is −4-4−4: 8−k22=−4.\frac{8-k^2}{2}=-4.28−k2​=−4. 8−k2=−88-k^2=-88−k2=−8 k2=16k^2=16k2=16 k=±4.k=\pm 4.k=±4.

  8. Now check the options. Among the given options, only k=−4k=-4k=−4 is present.

Therefore, the correct option is B.

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