Question

The perpendicular bisector of the line segment joining \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\] has \[y - \] intercept \[ - 4\]. Find the possible value of \[k\].
A. -4
B. 1
C. 2
D. -2

Answer 1

Hint: We will find the slope of the line that passes through the points \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\]. Apply the formula \[{m_1}{m_2} = - 1\], to get the slope of the perpendicular line. Then find the midpoint of the points \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\]. By using the midpoint and the slope of the perpendicular line we will find the equation of the line. From the equation of line, we will find the \[y - \]intercept and equate it with -4 to calculate the value of \[k\].

Formula used:
The slope of a line passes through the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].
The product of the slope of perpendicular lines is -1.
The coordinate of the midpoint of the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]
The equation of line passes through the point \[\left( {{x_1},{y_1}} \right)\] with slope \[m\] is \[y - {y_1} = m\left( {x - {x_1}} \right)\].

Complete step by step solution:
Given points are \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\].
Apply the slope formula on the given points
The slope of the line that passes through the points \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\] is \[{m_1} = \dfrac{{4 - 3}}{{1 - k}} = \dfrac{1}{{1 - k}}\].
Now we will calculate the slope of the line perpendicular to line that passes through the points \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\]:
So, apply the formula \[{m_1}{m_2} = - 1\].
\[\dfrac{1}{{1 - k}} \cdot {m_2} = - 1\]
\[ \Rightarrow {m_2} = - \left( {1 - k} \right)\]
\[ \Rightarrow {m_2} = - 1 + k\]
Now we will calculate the midpoint of the line segment joining by \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\].
Apply the formula \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\].
The coordinate of the midpoint is \[\left( {\dfrac{{1 + k}}{2},\dfrac{{4 + 3}}{2}} \right)\]
                                                          \[ = \left( {\dfrac{{1 + k}}{2},\dfrac{7}{2}} \right)\]
Now applying the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] to find the equation of perpendicular line
\[y - \dfrac{7}{2} = \left( { - 1 + k} \right)\left( {x - \dfrac{{1 + k}}{2}} \right)\]
\[ \Rightarrow y - \dfrac{7}{2} = - x + \dfrac{{1 + k}}{2} + xk - \dfrac{{k\left( {1 + k} \right)}}{2}\]
\[ \Rightarrow y = - x + xk + \dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2}\]
\[ \Rightarrow y = x\left( { - 1 + k} \right) + \dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2}\]
Now compare the equation \[y = x\left( { - 1 + k} \right) + \dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2}\] and \[y = mx + c\]
\[c = \dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2}\]
\[m = - 1 + k\]
So the \[y - \]intercepts is \[\dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2}\].
Equate the \[y - \]intercepts with -4.
\[\dfrac{{1 + k}}{2} - \dfrac{{k\left( {1 + k} \right)}}{2} + \dfrac{7}{2} = - 4\]
\[ \Rightarrow \dfrac{{1 + k}}{2} - \dfrac{{{k^2} + k}}{2} = - 4 - \dfrac{7}{2}\]
\[ \Rightarrow 1 + k - {k^2} - k = - 15\]
\[ \Rightarrow - {k^2} = - 16\]
\[ \Rightarrow {k^2} = 16\]
\[ \Rightarrow k = \pm 4\]
The possible value of \[k\] is -4.
Hence option A is correct.

Note: The equation line of perpendicular bisector of joining by the line segment \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[{\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2}\]. So, we can use this formula to find the equation of line perpendicular bisector of joining \[P\left( {1,4} \right)\] and \[Q\left( {k,3} \right)\].