Answer:
B. [tex]y=-\frac{1}{4}x-10[/tex]
Step-by-step explanation:
Given rectangle with sides J,K,L and M.
We need to find equation of sides that are perpendicular to side K.
For a rectangle adjacent sides are perpendicular to each other.
For side K, the adjacent sides are J and L. Hence, sides J and L are perpendicular to side L.
Finding equation of side J.
Points: (-5,4) and (3,2)
Slope of line [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
where [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] are points on the line.
Thus [tex]m=\frac{2-4}{3-(-5)}[/tex]
[tex]m=\frac{2-4}{3+5}[/tex]
[tex]m=\frac{-2}{8}[/tex]
Simplifying fraction.
[tex]m=-\frac{1}{4}[/tex]
Using point-slope equation to find equation of the line.
[tex](y-y_1)=m(x-x_1)[/tex]
Using point (-5,4)
[tex](y-4)=-\frac{1}{4}(x-(-5))[/tex]
[tex](y-4)=\-frac{1}{4}(x+5))[/tex]
Using distribution
[tex](y-4)=-\frac{1}{4}x-\frac{5}{4}[/tex]
Adding 4 to both sides.
[tex]y-4+4=-\frac{1}{4}x-\frac{5}{4}+4[/tex]
Taking LCD to add fraction.
[tex]y=-\frac{1}{4}x-\frac{5}{4}+\frac{16}{4}[/tex]
Equation of side J.
[tex]y=-\frac{1}{4}x-\frac{11}{4}[/tex]
Finding equation of side L.
Points: (-8,-8) and (0,-10)
Slope of line [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
where [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] are points on the line.
Thus [tex]m=\frac{-10-(-8)}{0-(-8)}[/tex]
[tex]m=\frac{-10+8}{0+8}[/tex]
[tex]m=\frac{-2}{8}[/tex]
Simplifying fraction.
[tex]m=-\frac{1}{4}[/tex]
Using point-slope equation to find equation of the line.
[tex](y-y_1)=m(x-x_1)[/tex]
Using point (0,-10)
[tex](y-(-10))=-\frac{1}{4}(x-0)[/tex]
[tex]y+10=-\frac{1}{4}x[/tex]
Subtracting both sides by 10.
[tex]y+10-10=-\frac{1}{4}x-10[/tex]
Equation of side L.
[tex]y=-\frac{1}{4}x-10[/tex]
The equation of side perpendicular to side K is represented by
[tex]y=-\frac{1}{4}x-10[/tex]
