Question

find the point of intersection of Yaxis and the perpendicular bisector of (2,-3) and (-4,1)

Answer 1

Let L1 be the line passing throught (2, -3) and (-4, 1)
Let L2 be the perpendicular bisector of L1

Gradient of L1 = (1- -3)/(-4 -2) = 4/-6 = -2/3
⇒ Gradient of L2 = 3/2 (The product of the gradients of 2 perpendicular lines = -1)

Mid-point of L1 = [ (2-4)/2 , (-3+1)/2] = (-1, -1)

∴ L2 has a gradient of = 3/2 and it passes through (-1, -1)

Equation of L2:

(y - -1)/ (x - -1) = 3/2
(y+1)/(x+1) = 3/2
y+1 = 3/2(x+1)
y+1 = 3/2x + 3/2

y = 3/2x +3/2 -1
y = 3/2x + 1/2

This equation is now in the form y = mc + c 
c is the y-intercept = 1/2

∴The point of intersection is (0, 1/2)

Answer 2

Answer:

The point of intersection of Y-axis and the perpendicular bisector of AB is $(0,\frac{1}{2})$.

Step-by-step explanation:

The given points are A(2,-3) and B(-4,1).

We have to the figure the equation of perpendicular bisector.

The mid point of given points is

$midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{2-4}{2},\frac{-3+1}{2})=(-1,-1)$

The perpendicular bisector of AB is perpendicular at point (-1,-1).

The slope AB is

$m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-3)}{-4-2}=\frac{4}{-6}=-\frac{2}{3}$

The product of slope of two perpendicular lines is -1. So the slope of perpendicular bisector of AB is $\frac{3}{2}$.

The equation of perpendicular bisector of AB is

$y-y-1=m(x-x_1)$

$y-(-1)=\frac{3}{2}(x-(-1))$

$y+1=\frac{3}{2}x+\frac{3}{2}$

$y=\frac{3}{2}x+\frac{1}{2}$                  ....(1)          

The equation of perpendicular bisector of AB is $y=\frac{3}{2}x+\frac{1}{2}$. To find the  point of intersection of Y-axis and the perpendicular bisector of AB, substitute x=0 in equation (1).

$y=\frac{3}{2}(0)+\frac{1}{2}$

$y=\frac{1}{2}$

Therefore the point of intersection of Y-axis and the perpendicular bisector of AB is $(0,\frac{1}{2})$.