If the perpendicular bisector of the line segment joining p(1,4) and q(k,3) h...

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Question Id : 11628 | Context :NIMCET 2024

If the perpendicular bisector of the line segment joining p(1,4) and q(k,3) has yintercept -4, then the possible values of k are

🎥 Video solution / Text Solution of this question is given below:

Step 1: Find midpoint of PQ

Midpoint = $\left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right)$

Step 2: Find slope of PQ

Slope of PQ = $\dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1}$

Step 3: Slope of perpendicular bisector = negative reciprocal = $k - 1$

Step 4: Use point-slope form for perpendicular bisector:

$y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right)$

Step 5: Find y-intercept (put $x = 0$ )

$y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right)$

$y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right)$

Given: y-intercept = -4, so:

$\dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4$

Multiply both sides by 2:

$7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8$

$\Rightarrow k^2 = 16 \Rightarrow k = \pm4$

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

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