Find the equation of the line thru (1-1) that is parallel to 3x+2y=1
There are two different ways on finding the equation of the line. Pick the one that is better for you:
First Way
First solve the equation $\displaystyle 3x+2y=1$ for y. You should get $\displaystyle y=-\frac{3}{2}x+\frac{1}{2}$. If the lines are parallel, then the new line with have the same slope. Since the new line must pass through (1,-1), we can substitute these values into the point-slope form of a line.
Point slope form: $\displaystyle y-y_1=m(x-x_1)$
$\displaystyle (y-(-1))=(-\frac{3}{2})(x-(1))\rightarrow y=-\frac{3}{2}x+\frac{3}{2}-1\rightarrow y=-\frac{3}{2}x+\frac{1}{2}$
Second Way
First solve the equation $\displaystyle 3x+2y=1$ for y. You should get $\displaystyle y=-\frac{3}{2}x+\frac{1}{2}$. If the lines are parallel, then the new line with have the same slope. We can substitute the value of the slope into the slope intercept form of a line:
Slope intercept form: $\displaystyle y=mx+b$
$\displaystyle y=-\frac{3}{2}x+b$.
To find b, plug in the point (1,-1)
$\displaystyle -1=-\frac{3}{2}(1)+b\rightarrow -1+\frac{3}{2}=b\rightarrow b=\frac{1}{2}$.
Thus the equation of the line is $\displaystyle y=-\frac{3}{2}x+\frac{1}{2}$.
Hope this helped you out!