Find the equation of the line through point (−1,2) and perpendicular to x+3y=3

Answer 1

Perpendicular means the slope of the line we are seeking is the negative reciprocal of the slope of x + 3y = 3.

To find the slope of x + 3y = 3, put it in slope-intercept form (y = mx + b). Then the slope will be the coefficient in front of the "x".

x + 3y = 3

3y = - x + 3

y = (-1/3)x + 1

So the slope is -1/3. The negative reciprocal to -1/3 is +3/1 or 3

So the line we are seeking has a slope of 3 and goes through the point (-1, 2). We can use the point-slope form of a line to write the equation. The point-slope form is:

y - y₁ = m(x - x₁)

Where m = 3 and (x₁, y₁) is (-1, 2)

y - 2 = 3(x - -1)

y - 2 = 3(x + 1)

Your question does not specify what form you need to give your answer in so you can leave it like this if you want. Or, if you want, you can multiply it out and combine like terms:

y - 2 = 3x + 3

y = 3x + 5

Or you can even manipulate it to get it in standard form:

-5 = 3x - y

3x - y = -5