Question

Write the equation of a line passing through (6,-2) and perpendicular to -2x + 3y= -6

Answer 1

Perpendicular lines refers to a pair of straight lines that intercept each other. The slopes of this lines are opposite reciprocal, meaning that it's multiplication is -1.
On this case they give you the equation of a line and a point, and is asked to find the equation of a line that is perpendicular to the given one, and that passes through this point.


-2x+3y=-6                 Add 2x in both sides
3y=2x-6                    Divide by 3 in both sides to isolate y
y=2/3x-6/3


The slope of the given line is 2/3, which means that the slope of a line perpendicular to this one, needs to be -3/2. Now you need to find the value of b or the y-intercept by substituting the given point into the formula y=mx+b, where letter m represents the slope.

y=mx+b                 Substitute the given point and the previous slope found
-2=(-3/2)(6)+b       Combine like terms
-2=-9+b                 Add 9 in both sides to isolate b
7=b

The equation that represents the line perpendicular to -2x+3y=-6 and that passes through the point (6,-2), is y=-3/2x+7.