JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 9)
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 :
$$\sqrt {14} $$
-4
–2
$$\sqrt {15} $$
Explanation
$${m_{PQ}} = {{4 - 3} \over {1 - k}} $$
$$ \therefore $$ 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
$$y = {7 \over 2} - \left( {{{{k^2} - 1} \over 2}} \right) = - 4$$
$${{{k^2} - 1} \over 2} = {{15} \over 2} \Rightarrow k = \pm 4$$
$$ \therefore $$ 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
$$y = {7 \over 2} - \left( {{{{k^2} - 1} \over 2}} \right) = - 4$$
$${{{k^2} - 1} \over 2} = {{15} \over 2} \Rightarrow k = \pm 4$$
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