parallel and perpendicular lines

Question

identify an equation in point-slope form for the line perpendicular to y=- 1/3 x-6 that passes through (-1,5)

Answer 1

For the given equation. the slope is -1/3. A line parallel to it will have a slope of +3 (the negative inverse)

You are asked for the new line in the "point-slope" form, which is (y - a) = m*(x - b), where a and b are the y and x offsets.

Plug in m and the given offsets to get:

(y - 5) = 3*(x +1 )

Answer 2

Hi Unique

Remember that point-slope form looks like y-y₁=m(x-x₁), passes through (x₁,y₁) and m is the slope.

Remember also that perpendicular lines have opposite reciprocal slopes, so if one has a slope of a/b, the other has a slope of -b/a.

If we want a line perpendicular to y=- 1/3 x-6, then we want the slopes to be opposite reciprocal. This first line is written in slope-intercept form (y=mx+b), so the slope is -1/3. The opposite reciprocal of -1/3 is 3.

We need the equation for a line with a slope of 3 that passes through (-1,5), so y-5=3(x-(-1)). Simplified, it will look like y-5=3(x+1), but don't do anything else, or it won't be consider point-slope form any more!

Answer 3

To find a perpendicular line, the slope is going to be the negative reciprocal to the original equation:

-1/3 → 3

From there, we use the point-slope formula:

y - y₁ = m(x - x₁)

y - 5 = 3(x + 1)