Question

Find the point of intersection of y-axis and the perpendicular bisector of $(2, - 3)$ and $( - 4,1)$.

Answer 1

Hint: As in this question first try to find out the midpoint of points $(2, - 3)$ and $( - 4,1)$ then for the slope as we know that ${m_1}.{m_2} = - 1$ if line is perpendicular hence one slope we know other will find out then find out the equation of line perpendicular to it and at last find the intersection with y-axis

Complete step-by-step answer:
As in the given question first we have to find the equation of perpendicular bisector of points $(2, - 3)$ and $( - 4,1)$
Perpendicular bisector means line passing through midpoint and perpendicular to the point $(2, - 3)$ and $( - 4,1)$
               

Hence the midpoint of point $(2, - 3)$ and $( - 4,1)$ is M equal to
$x = \dfrac{{{x_1} + {x_2}}}{2}$ and $y = \dfrac{{{y_1} + {y_2}}}{2}$
or
$x = \dfrac{{2 + ( - 4)}}{2}$ and $y = \dfrac{{ - 3 + 1}}{2}$
$x = - 1$ and $y = - 1$
Hence the midpoint of A $(2, - 3)$ and B $( - 4,1)$ is M $( - 1, - 1)$
Now for the equation of line we need the slope of the line , As we know that the line MC is perpendicular to the line AB joining point $(2, - 3)$ and $( - 4,1)$
Hence we know that ${m_1}.{m_2} = - 1$
means Slope of line AB $ \times $ Slope of MC = $ - 1$
Slope of AB equal to = ${m_1}$ = $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
= $\dfrac{{1 - ( - 3)}}{{ - 4 - 2}}$
= $\dfrac{{1 + 3}}{{ - 4 - 2}}$
= $\dfrac{4}{{ - 6}}$
${m_1}$ = $ - \dfrac{2}{3}$
Slope of MC = ${m_2}$
Hence for ${m_2} = -\dfrac{1}{{{m_1}}}$= $\dfrac{1}{{ - \dfrac{2}{3}}}$
${m_2} = \dfrac{3}{2}$
Equation of line MC is whose one point is $( - 1, - 1)$ and slope is ${m_2} = \dfrac{3}{2}$
$(y - {y_1}) = {m_2}(x - {x_1})$
$(y + 1) = -\dfrac{3}{2}(x + 1)$
In the question we have to find the point of intersection of this line with the y-axis .
For the y-axis $X = 0$ ,put it in the equation of line MC
$(y + 1) = \dfrac{3}{2}(0 + 1)$
$y + 1 = \dfrac{3}{2}$
$y = \dfrac{3}{2} - 1$
$y = \dfrac{1}{2}$
Hence the point of intersection of y-axis and the perpendicular bisector of $(2, - 3)$ and $( - 4,1)$ is $\left( {0, \dfrac{1}{2}} \right)$

Note: As in the question we use for the midpoint of line AB is $x = \dfrac{{{x_1} + {x_2}}}{2}$ and $y = \dfrac{{{y_1} + {y_2}}}{2}$
 this is nothing but the section formula $x = \dfrac{{m{x_1} + n{x_2}}}{{m + n}}$ or $y = \dfrac{{m{y_1} + n{y_2}}}{{m + n}}$ which have $m = 1,n = 1$ because we have to find midpoint M hence M divides line AB in equal ratio .
For finding the equation of line we need one point and the slope of that line as in this question we find a point as midpoint M and the slope with formula ${m_1}.{m_2} = - 1$