Question

find the equation for the line that passes through the point (2,-2) and that is parallel to the line with the equation 3/2x-2y=23/2

Answer 1

First let's rearrange the equation so it's in y = mx +b form.\[-2y=\frac{ 23 }{ 2 }-\frac{ 3 }{ 2 }x \rightarrow y =\frac{ \frac{ 23 }{ 2 }-\frac{ 3 }{ 2 }x }{ -2 }\rightarrow y=\frac{ 3 }{ 4 }x-\frac{ 23 }{ 4 }\]Now since we we know that the line is parallel, it must have the same slope. Therefore, the equation of the parallel line must be in the form:\[y=\frac{ 3 }{ 2 }x+b\]We need to solve for b for this parallel line and we can do that by using the point we are already given. The point we are given is (2, -2). We plug 2 from this point for x and -2 for y, and then solve for b:\[-2=\frac{ 3 }{ 2 }(2)+b \rightarrow -2 = 3+b \rightarrow b = -5\]So now we know what the value of b is in the equation of our parallel line. Therefore, the equation of the parallel line going throw (2, -2) is:\[\y=\frac{ 3 }{ 2 }x-5\]