JEE MAIN - Mathematics (2008 - No. 18)
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 :
1
2
-2
-4
Explanation
Slope of $$PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$$
$$\therefore$$ Slope of perpendicular bisector of
$$PQ = \left( {k - 1} \right)$$
Also mid point of
$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$
Equation of perpendicular bisector is
$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$
$$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$
$$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$
$$\therefore$$ $$y$$-intercept $$ = {{8 - {k^2}} \over { - 2}} = - 4$$
$$ \Rightarrow $$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$
$$\therefore$$ Slope of perpendicular bisector of
$$PQ = \left( {k - 1} \right)$$
Also mid point of
$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$
Equation of perpendicular bisector is
$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$
$$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$
$$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$
$$\therefore$$ $$y$$-intercept $$ = {{8 - {k^2}} \over { - 2}} = - 4$$
$$ \Rightarrow $$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$
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