Question

find the equation of the line perpendicular to 3x-7y=42 which passes through the point (-3,-8)

Answer 1

For perpendicular lines, m1m2 = -1 or m2 = -1/m1; where m1 and m2 are the slopes of the lines.
Here line 1 is 3x - 7y = 42
7y = 3x - 42
y = 3/7 x - 6; Hence m1 = 3/7
m2 = -1/(3/7) = -7/3
Required equation y - y1 = m2(x - x1)
y - (-8) = -7/3(x - (-3))
y + 8 = -7/3(x + 3)
y + 8 = -7/3 x - 7
y = -7/3 x - 7 - 8
y = -7/3 x - 15