147k views
4 votes
Find the equation of a line that is perpendicular to y = 3x + 3 and passes through the point (-3, -1).

a) y = -1/3x - 2
b) y = -1/3x + 8
c) y = 3x - 2
d) y = 3x + 8

User Nikoshr
by
7.6k points

1 Answer

5 votes

Final answer:

To find the perpendicular line equation, calculate the negative reciprocal of the original line's slope and use the given point to determine the y-intercept. The correct equation is y = -1/3x - 2, option a.

Step-by-step explanation:

The question asks for the equation of a line that is perpendicular to the line represented by the equation y = 3x + 3 and that passes through the point (-3, -1). Since the slope of the given line is 3 (the coefficient of x), a line perpendicular to it will have a slope that is the negative reciprocal of 3, which is -1/3. Using the slope-intercept form y = mx + b, and inserting the point (-3, -1), we get y = (-1/3)x + b. Now we solve for b using the coordinates of the point.

Substituting x = -3 and y = -1 into the equation gives us -1 = (-1/3)(-3) + b, which simplifies to -1 = 1 + b. Subtracting 1 from both sides, we find b = -2. Therefore, the equation of our desired line is y = -1/3x - 2, which corresponds to option a.

User Medilies
by
7.0k points