In slope-intercept form, what is the equation of a line perpendicular to y = 3x– 5 that passes through the point (-3, -6)? 1. y = 3x – 7 2. y = 1/3x + 7 3. y = -1/3x – 7 4. y = -3x + 7

Answer is 3

So we have a slope m = -1/3   and a point (-3,-6)  so use the point slope formula (Y-Y1) = m(X-X1)In this case (X1,Y1) = (-3,-6) (Y -( -6) = (-1/3)(X-(-3) Y+6 = -1/3X-1 Y = -1/3X-7 would be the equation of the perpendicular line through (-3,-6)

y = -(1/3) x - 7

Explanation:
The equation of the linear line has the following general form:
y = mx + c
where m is the slope and c is the y-intercept

1- getting the slope:
We are given that the line we are looking for is parallel to the line:
y = 3x - 5
This means that:
m of required line * m of given line = -1
m of required line * 3 = -1
m of required line = -1/3

2- getting the value of c:
To get the value of c, we will simply use the given point (-3,-6) to substitute in the equation and solve for c as follows:
y = -(1/3) x + c
-6 = -(1/3)(-3) + c
-6 = 1 + c
c = -7

Based on the above, the equation of the line is:
y = -(1/3) x - 7

Hope this helps :)