Question

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

• A. 1
• B. 2
• C. -2
• D. -4

Answer 1

Answer: (D)

-4

Answer 2

Slope of $PQ = \frac{3-4}{k-1} = \frac{-1}{k-1}$ $\therefore$ Slope of perpendicular bisector of $PQ =\left(k-1\right)$ Also mid point of $PQ \left(\frac{k+1}{2}, \frac{7}{2}\right)$ $\therefore$ Equation of perpendicular bisector is $y - \frac{7}{2} = \left(k-1\right) \left(x - \frac{k+1}{2}\right)$ $\Rightarrow 2y -7 = 2\left(k-1\right)x -\left(k^{2 } - 1\right)$ $\Rightarrow 2\left(k-1\right)x - 2y + \left(8 - k^{2}\right) = 0$ $\therefore$ y-intercept $= - \frac{8-k^{2}}{-2} = - 4$ $\Rightarrow 8 -k^{2} = - 8$ or $k^{2} = 16 \Rightarrow k = \pm4$