Question

find the equation of the line that goes through (8,-1) and is perpendicular to y-2=4/3(x+9).

Answer 1

point (8, -1)

y - 2 = 4/3(x + 9)

y = 4/3(x + 9) + 2

y = 4/3 * x + 4/3 * 9 + 2

y = 4/3x + 12 + 2

y = 4/3x + 14

from y = mx + c

slope = m

slope = 4/3

since it is perpendicular to the point

m1 x m2 = -1

m1 = 4/3

4/3 x m2 = -1

m2 = -1 / 4/3

m2 = -3/4

our new slope = -3/4

from (y - y1) = m(x - x1)

y1 = -1

x = 8

(y- (-1)) = -3/4(x - 8)

y + 1 = -3/4(x - 8)

y = -3/4(x-8) - 1

the new equation is -3/4(x - 8) - 1