Question 225 of 303MediumMCQ
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: Solutions · 1
Official solution
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−k2=−8 or \[{k^2} = 16 \Rightarrow k = \pm 4\] Answer the question to unlock