On a coordinate plane, a line goes through (negative 3, 2) and (2, negative 1). A point is at (3, 0). What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)? 3x + 5y = −9 3x + 5y = 9 5x − 3y = −15 5x − 3y = 15

5x -3y = 15

Step-by-step explanation:

First find the slope of the line through ( -3,2) and (2,-1)

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

   =( -1-2)/(2--3)

   = -3/(2+3)

    = -3/5

We want a line that is perpendicular so the slopes will multiply to -1

m * -3/5 = -1

Multiply each side by -5/3

m * -3/5 * -5/3 = -1 *-5/3

m = 5/3

The slope of the perpendicular line is 5/3

We have a slope and a point

Using point slope form

y-y1 = m(x-x1)

y-0 = 5/3( x-3)

y = 5/3( x-3)

Distribute

y = 5/3x -5

Multiply each side by 3

3*y = 3(5/3x -5)

3y = 5x - 15

Subtract 5x from each side

-5x+3y = -15

Multiply each side by -1

5x -3y = 15

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8