Question

write the equation of a line that is perpendicular to the line y=-6 and passes through the point (3,7)

Answer 1

Step-by-step explanation:

when two lines are perpendicular to each other( the product of their slope; m1m2 = -1)

m1 = -1/m2

m1 = -1/-6= 1/6

using the formula m= y-y1 / x - x1

1/6 = y -7 /x - 3

6(y -7) = x -3

6y -42 = x - 3

6y - x -39 =0

Divide both sides by 6

y - 1/6 x - 13/2= 0

y = 1/6 x + 13/2