Question

Write an equation of the line that passes through (18, 2) and is parallel to the line 3y−x=−123y−x=−12.

Answer 1

Find the slope of the given line, 3y−x=−12: Solving this equation for y, we get

3y = x - 12, so that y = (1/3)x - 4. This slope is 1/3. The slope of a line perpendicular to the given line is the neg. reciprocal of 1/3, which is -3.

In point-slope form, the equation of the new (perpendicular) line is

y - 2 = -3(x-18).