Question

JK is formed by J(-12, 3) and K(8,-5). If
line r is the perpendicular bisector of JK, write
an equation for r in slope-intercept form??

Answer 1

Given:

JK is formed by J(-12, 3) and K(8,-5).

Line r is the perpendicular bisector of JK.

To find:

The equation of line r is slope intercept form.

Solution:

Line r is the perpendicular bisector of JK. So, it passes through the midpoint of J and K.

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-12+8}{2},\dfrac{3+(-5)}{2}\right)$

$Midpoint=\left(\dfrac{-4}{2},\dfrac{-2}{2}\right)$

$Midpoint=\left(-2,-1\right)$

Slope of JK is

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

$m=\dfrac{-5-3}{8-(-12)}$

$m=\dfrac{-8}{20}$

$m=\dfrac{-2}{5}$

Product of slopes of two perpendicular lines is -1.

$m_1\times m_2=-1$

$-\dfrac{2}{5}\times m_2=-1$

$m_2=\dfrac{5}{2}$

The slope of perpendicular line r is $\dfrac{5}{2}$ and it passes through (-2,-1). So, the equation of perpendicular line is

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

$y-(-1)=\dfrac{5}{2}(x-(-2))$

$y+1=\dfrac{5}{2}(x+2)$

$y+1=\dfrac{5}{2}x+5$

$y=\dfrac{5}{2}x+5-1$

$y=\dfrac{5}{2}x+4$

Therefore, the equation of line r is $y=\dfrac{5}{2}x+4$.

Answer 2

The required equation of line "r" in slope-intercept form is y = 5/2x + 15

First, we need to find the equation of a line in slope-intercept form as

y = mx + b

m is the slope of the line

b is the y-intercept

Given the coordinate points  J(-12, 3) and K(8,-5)

Get the slope of the line

Slope  = -5-3/8+12

Slope = -8/20

Slope - -2/5

Slope of the line "r" perpendicular is 5/2

Get the y-intercept of line r

Substitute (8, -5) and m = 5/2 into y = mx + b

-5 = -5/2(8) + b

-5 = -20 + b

b = 15

Hence the required equation of line "r" in slope-intercept form is y = 5/2x + 15

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