On a coordinate plane, a line goes through (negative 8, 10), (0, 0), and (8, negative 10). What is the equation of the line that is perpendicular to the given line and passes through the point (5, 3)? 4x – 5y = 5 5x + 4y = 37 4x + 5y = 5 5x – 4y = 8

4x-5y =5

Step-by-step explanation:

First we need to find the slope of the line

m = (y2-y1)/(x2-x1)

   = (0-10)/(0- -8)

   = -10/(0+8)

   -10/8  

   -5/4

We want a line with a perpendicular slope

That line will have a negative reciprocal slope

-1 (-5/4)

4/5

We have the slope and a point

y= mx+b is the slope intercept form of a line

y = 4/5x+b

Substitute the point into the equation

3 = 4/5(5)+b

3 =4+b

b = -1

The equation of the line is y = 4/5x-1

Multiply each  side by 5 to get rid of the fractions

5y = 4x -5

Subtract 4x from each side

-4x+5y = -5

Multiply each side by -1

4x-5y =5

4x - 5y = 5

Step-by-step explanation:

Slope of the given line:

(-10-0)/(8-0)

-10/8

-5/4

Slope of the Perpendicular line

4/5

y = ⅘x + c

3 = (4/5)(5) + c

3 = 4 + c

c = -1

y = (4/5)x - 1

5y = 4x - 5

4x - 5y = 5

its c or d   take a lucky guess my dude

y + 11 = (1/4)(x + 16)

step-by-step explanation: