Question

The equation of perpendicular bisector of the line segment joining the points $\left( {1,2} \right)$ and $\left( { - 2,0} \right)$ is
a)$5x + 2y = 1$ b.) $4x + 6y = 1$
c)$6x + 4y = 1$ d.) none

Answer 1

Hint:- Equation of the perpendicular bisector can be found by using slope of the line. So at first, try to find out the slope of AB and hence the slope of CD.

Complete step-by-step answer:
In the above question, the equation of the perpendicular bisector of the line segment is asked to us.
The two given points are $\left( {1,2} \right)$ and $\left( { - 2,0} \right)$

Let ‘p’ be the mid-point of the line AB joining the points A(1,2) and B(-2,0)

So, using midpoint formula:-
${\text{P}} \equiv \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Here, ${x_1} = 1\;\;;{y_1} = 2$
${x_2} = - 2\;;{y_2} = 0$
So, ${\text{P}} \equiv \left( {\dfrac{{1 + \left( { - 2} \right)}}{2},\dfrac{{2 + 0}}{2}} \right)$
$\Rightarrow {\text{P}} \equiv \left( {\dfrac{{1 - 2}}{2},\dfrac{2}{2}} \right)$
$\Rightarrow {\text{P}} \equiv \left( {\dfrac{{ - 1}}{2},1} \right) \to$ midpoint of AB.
Now, slope of AB $= \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
So, ${x_1} = 1\;;{y_1} = 2$
${x_2} = - 2\,;{y_2} = 0$
⇒ slope of AB $= \dfrac{{0 - 2}}{{ - 2 - 1}} = \dfrac{{ - 2}}{{ - 3}}$
$\therefore$ slope of AB = {\raise0.5ex\hbox{\scriptstyle 2} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}.
So, slope of line perpendicular to AB
$= \dfrac{{ - 1}}{{{\text{slope}}\;{\text{of}}\;{\text{AB}}}}$
= \dfrac{{ - 1}}{{{\raise0.5ex\hbox{\scriptstyle 2} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}}} = \boxed{\dfrac{{ - 3}}{2}} \to Slope of perpendicular line to AB.
$\therefore$ equation of perpendicular bisector;
$\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Where $\left( {{x_1},{y_1}} \right)$ are the points and ‘m’ is the slope;
\Rightarrow y - 1 = \dfrac{{ - 3}}{2}\left( {x - \left( {{\raise0.5ex\hbox{\scriptstyle { - 1}} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} \right)} \right)
$\Rightarrow y - 1 = \dfrac{{ - 3}}{2}\left( {x + \dfrac{1}{2}} \right)$
$\Rightarrow 2y - 2 = - 3\left( {x + \dfrac{1}{2}} \right)$
$\Rightarrow 2y - 2 = - 3\left( {\dfrac{{2x + 1}}{2}} \right)$
$\Rightarrow 4y - 4 = - 6x - 3$
$\Rightarrow 4y + 6x = - 3 + 4$
$\Rightarrow 6x + 4y = 1$
$\Rightarrow \boxed{6x + 4y - 1 = 0}$
$\therefore$ equation of line $6x + 4y - 1 = 0$
Note:- For the equation of perpendicular bisectors, we need the points, and for that purpose we find the midpoint of line AB.
Midpoint of a line with points ${\text{A}}\left( {{x_1},{y_1}} \right)$ and ${\text{B}}\left( {{x_1},{y_2}} \right)$ $\equiv \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
and $e{q^n}$of line:-$\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$ where $\left( {{x_1},{y_1}} \right)$ are the points and ‘m’ is the slope of the line.