Acircle with center c(4,-2) contains the point d(8,1). what is the equationi of the line perpendicular to the radius of the circle passing through point c?

  y = (-4/3)(x -4) -2

Step-by-step explanation:

The slope of the radius to point D is ...

  m = (change in y)/(change in x) = (1-(-2))/(8-4) = 3/4

The slope of the perpendicular line will be the negative reciprocal of this:

  m' = -1/m = -1/(3/4) = -4/3

The point-slope form of the equation for a line with slope m through point (h, k) can be written as ...

  y = m(x -h) +k

We want a line with slope m' through point C:

  y = (-4/3)(x -4) -2

omg what happend to your question? if you could edit this question i can

step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length