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QJEEM20Ma586c: JEE Main 2020 Mathematics - Class 11 JEEM20Ma586c Solution: If the perpendicular bisector of the line segment joining the…

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 :

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{m_{PQ}} = {{4 - 3} \over {1 - k}}

\therefore

Slope of perpendicular bisector of PQ,

{m_ \bot } = k - 1

mid point of PQ =

\left( {{{k + 1} \over 2},{7 \over 2}} \right)

equation of perpendicular bisector

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

for y intercept put x = 0

y = {7 \over 2} - \left( {{{{k^2} - 1} \over 2}} \right) = - 4

{{{k^2} - 1} \over 2} = {{15} \over 2} \Rightarrow k = \pm 4