Question

Line segment Xy has endpoints x(5,7) and y(-3,3). Find the equation for the perpendicular bisector of line segment xy

Answer 1

Answer: The equation for the perpendicular bisector of line segment xy is $y=-2x+7$.

Explanation:

It is given that the line segment xy has endpoints x(5,7) and y(-3,3).

The bisector divides the line segment xy in two equal parts, so the bisector must be passing through the midpoint of xy.

Midpoint of two points (x_1,y_1) and (x_2,y_2) is calculated as,

$\text{Midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Midpoint of xy is,

$\text{Midpoint}=(\frac{5-3}{2},\frac{7+3}{2})=(1,5)$

So, the perpendicular bisector must be passing through the point (1,5).

The slope of line passing through the point (x_1,y_1) and (x_2,y_2) is calculated as,

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{3-7}{-3-5}=\frac{1}{2}$

Slope of line segment is $\frac{1}{2}$. Therefore the slope of perpendicular bisector is -2 because the the product of slopes of two perpendicular lines is always -1.

The line passing through the point (x_1,y_1) with slope m is defined as,

$y-y_1=m(x-x_1)$

Bisector passing through the point (1,5) with slope 2.

$y-5=-2(x-1)$

$y=-2x+2+5$

$y=-2x+7$

Therefore, the equation for the perpendicular bisector of line segment xy is $y=-2x+7$.