Finding perpendicular bisector of the line segement joining $ (-1,4)\;\text{and}\;(3,-2)$ 
Find the perpendicular bisector of the line joining the points $(-1,4)\;\text{and}\;(3,-2).\;$

I know this is a very easy question, and the answer is an equation. So any hints would be very nice. thanks
 A: Hints: 


*

*find the midpoint $p$ of the line segment connecting two points $(-1,4)$, $(3,-2)$
$$\text{midpoint} = (x_p, y_p)\; = \left(\dfrac{x_2 + x_1}{2}, \dfrac{y_2 + y_1}{2}\right)$$

*find slope $m_1$ of the line $\mathcal{l}_1$ connecting $(-1,4)$, $(3,-2)$
$$m_1 = \dfrac{y_2 - y_1}{x_2 - x_1}$$

*The line perpendicular to $\mathcal{l}_1$ will have slope $m_2 = \dfrac{-1}{m_1}$ 
Use the "point-slope" formula to obtain the equation of the desired line (the perpendicular bisector), using point $p = (x_p, y_p)$ from $(1)$ and slope $m_2$ from $(3)$: $$y - y_p = m_2(x - x_p)\tag{point-slope form}$$
A: If the point $P=(x,y)$ lies on the perpendicular bisector of the points $A=(-1,4)$ and 
$B= (3, -2)$, then the distances $PA$ and $PB$ must be the same.
Then
$$
PA^2 = PB^2  \\
\Rightarrow (x+1)^2 + (y-4)^2 = (x-3)^2 + (y+2)^2 \\
\Rightarrow  x^2 +2x + 1 + y^2 -8y +16 = x^2 -6x +9 +y^2 + 4y + 4 \\
\Rightarrow 2x - 8y +17 = -6x +4y +13 \\
\Rightarrow 8x - 12y + 4 = 0 \\
\Rightarrow 2x -3y + 1 = 0
$$
So, the perpendicular bisector has equation $2x -3y + 1 = 0$.
A: Hint: 
(a) find the midpoint of the two points $(-1,4)$, $(3,-2)$
(b) find gradient of line joining $(-1,4)$, $(3,-2)$
(c) using the gradient, find the normal
A: Hint: The line must be orthogonal to the difference vector $(3-(-1),-2-4)$ and pass through the midpoint $(\frac{-1+3}2,\frac{4-2}2)$.
A: if $y = mx + c$ is a line equation, then equation for perpendicular lines to it will be like $y = (-\frac{1}{m}x + c')$ 
