Answer:
[tex]y=\frac{-4}{3}x+\frac{10}{3}[/tex]
Step-by-step explanation:
A circle with center C (4,-2) contains the point D (8,1).
Lets find out the slope of the line that contains point C and D
Slope = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
C (4,-2) is (x1,y1) and D (8,1) is (X2,y2)
Slope = [tex]\frac{1+2}{8-4}=\frac{3}{4}[/tex]
To get slope of perpendicular line we take negative reciprocal of 3/4 that is -4/3
the line passes throught point C
m= -4/3 , point (4,-2)
Use point slope formula
y-y1=m(x-x1) x1= 4, y1= -2, plug in all the values
[tex]y+2=\frac{-4}{3}(x-4)[/tex]
[tex]y+2=\frac{-4}{3}x+\frac{16}{3}[/tex]
Subtract 2 on both sides
[tex]y=\frac{-4}{3}x+\frac{10}{3}[/tex]
