Question:
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 :
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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The perpendicular bisector passes through the midpoint of PQ with slope perpendicular to PQ; set its y-intercept to -4.
Updated On: Nov 5, 2026
- \(\sqrt{14}\)
- -4
- -2
- \(\sqrt{15}\)
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The Correct Option is B
Solution and Explanation
Midpoint of PQ: \(M = \left(\frac{1+k}{2}, \frac72\right)\). Slope of PQ \(= \frac{3-4}{k-1} = -\frac{1}{k-1}\), so the perpendicular bisector has slope \(k-1\).
Its equation: \(y - \frac72 = (k-1)\left(x - \frac{1+k}{2}\right)\).
At \(x = 0\): \(y = \frac72 - \frac{(k-1)(k+1)}{2} = \frac{7 - k^2 + 1}{2} = \frac{8-k^2}{2}\).
Set \(\frac{8-k^2}{2} = -4 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4\).
Only \(-4\) is among the options.
Answer: (2) -4
Its equation: \(y - \frac72 = (k-1)\left(x - \frac{1+k}{2}\right)\).
At \(x = 0\): \(y = \frac72 - \frac{(k-1)(k+1)}{2} = \frac{7 - k^2 + 1}{2} = \frac{8-k^2}{2}\).
Set \(\frac{8-k^2}{2} = -4 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4\).
Only \(-4\) is among the options.
Answer: (2) -4
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