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

Updated On: Jul 7, 2022
  • 1
  • 2
  • -2
  • -4
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The Correct Option is D

Solution and Explanation

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$
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Concepts Used:

Coplanarity of Two Lines

Condition for Coplanarity Using Cartesian Form

The condition for coplanarity in the Cartesian form appears from the vector form.

Let's consider two points L (a1, b1, c1) & Q (a2, b2, c2) in the Cartesian plane,

Presuppose that there are two vectors q1 and q2. Their direction ratios are subjected by {x1, y1, z1}, and {x2, y2, z2} respectively.

The vector form of equation of the line in connection to L and Q can be stated as under:

LQ = (a2 – a1)i + (b2 – b1)j + (c2 – c1)k

Q1 = x1i + y1j + z1k

Q2 = x2i + y2j + z2k

Condition for Coplanarity Using Vector Form

For the derivation of the condition for coplanarity in vector form, we shall take into consideration the equations of two straight lines to be as stated below:

r1 = l1 + λq1

r2 = l2 + λq2

The condition for coplanarity in vector form is that the line in connection to the two points should be perpendicular to the product of the two vectors, q1 and q2.