equation of a line through (0,6) and perpendicular to 35x-7y=-7

Question

find the equation of a line through (0,6) and perpendicular to 35x-7y=-7

Answer 1

Line equation is 35x-7y = -7

Subtract 35x from each side.

35x-35x-7y = -7-35x

-7y = -35x-7

Divide to each side by negitive 7.

-7y/-7 = (-35/-7)x-7/-7

y = 5x+1

Slope is say m1 = 5.

Required line slope is be say m2 = -1/5

Point say(x1,y1) = (0,6)

Equation is y-6 = 1/-5(x-0)

y-6 = (-1/5)x

Add 6 to each side.

y-6+6 = (-1/5)x+6

Required line equation is y = (-1/5)x+6.