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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

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5 Answers

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Hints:

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

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

3. 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 $(3)$: $$y - y_p = m_2(x - x_p)\tag{point-slope form}$$

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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 $2x -3y + 1 = 0$.

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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

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Hint: The line must be orthogonal to the difference vector $(3-(-1),-2-4)$ and pass through the midpoint $(\frac{-1+3}2,\frac{4-2}2)$.

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if $y = mx + c$ is a line equation, then equation for perpendicular lines to it will be like $y = (-\frac{1}{m}x + c')$

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