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QJEEM08Ma94e2: JEE Main 2008 Mathematics - Class 11 JEEM08Ma94e2 Solution: The perpendicular bisector of the line segment joining P(1, 4)…

The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is :

Multiple Choice Options

  • A 1
  • B 2
  • C -2
  • D -4

Step-by-Step Verified Solution

Slope of PQ=3−4k−1=−1k−1PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}

∴\therefore Slope of perpendicular bisector of

PQ=(k−1)PQ = \left( {k - 1} \right)

Also mid point of

PQ(k+12,72).PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).

Equation of perpendicular bisector is

y−72=(k−1)(x−k+12)y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)

⇒2y−7=2(k−1)x−(k2−1)\Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)

⇒2(k−1)x−2y+(8−k2)=0\Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0

∴\therefore yy -intercept =8−k2−2=−4= {{8 - {k^2}} \over { - 2}} = - 4

⇒\Rightarrow 8−k2=−88 - {k^2} = - 8 or k2=16⇒k=±4{k^2} = 16 \Rightarrow k = \pm 4

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