Question

Equation of the perpendicular bisector of the line segment joining the points $(7,4)$ and $(-1,-2)$ is
A. $4x-3y=15$
B. $3x+4y=15$
C. $4x+3y=15$
D. None of these

Answer 1

Hint: In this question, we are to find the equation of the perpendicular bisector of the line segment. For finding this, the mid-point of the given points is calculated and then the slope of the line is calculated. With this two, we can easily calculate the equation of the line.

Formula Used:The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
We can also write the equation of the line in point-slope form as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
The slope of the line with the equation $ax+by+c=0$ is $m=\frac{-a}{b}$
Since the product of the slopes of the perpendicular lines is $-1$, the slope of the perpendicular line becomes $\frac{-1}{m}$.
The equation of the line that is perpendicular to the line with the equation $ax+by+c=0$ and passing through a point $({{x}_{1}},{{y}_{1}})$ is $b(x-{{x}_{1}})-a(y-{{y}_{1}})=0$



Complete step by step solution:Given that, a perpendicular bisector bisects the line segment joining the points $(7,4)$ and $(-1,-2)$.
So, the required line passes through the mid-point of the given points.
I.e.,
$\begin{align}
  & M=\left( \frac{7+(-1)}{2},\frac{4+(-2)}{2} \right) \\
 & \text{ }=\left( \frac{6}{2},\frac{2}{2} \right) \\
 & \text{ }=\left( 3,1 \right) \\
\end{align}$
The slope of the given line is
$\begin{align}
  & m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\frac{-2-4}{-1-7} \\
 & \text{ }=\frac{6}{8} \\
 & \text{ }=\frac{3}{4} \\
\end{align}$
Therefore, the slope of the required perpendicular bisector is
$\begin{align}
  & =\frac{-1}{m} \\
 & =\frac{-1}{\frac{3}{4}} \\
 & =\frac{-4}{3} \\
\end{align}$
Thus, the equation of the line with the slope $m=\frac{-4}{3}$ and passing through the point $({{x}_{1}},{{y}_{1}})=(3,1)$ is,
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & y-1=\frac{-4}{3}(x-3) \\
 & 3y-3=-4x+12 \\
 & \Rightarrow 4x+3y=15 \\
\end{align}$



Option ‘C’ is correct

Note: Here we can use the point-slope form to find the equation of the required perpendicular bisector. The slope can be calculated by the condition that the product of slopes of perpendicular lines is $-1$. By applying appropriate formulae, the required values are calculated