## College Algebra (6th Edition)

Point-slope form: $y-2=\dfrac{2}{3}(x+2)$ General form: $2x-3y+10=0$
Passing through $(-2,2)$ and parallel to the line whose equation is $2x-3y-7=0$ Rewrite the given line's equation in slope-intercept form. Start by taking $3y$ to the right side: $2x-3y-7=0$ $2x-7=3y$ Rearrange: $3y=2x-7$ Take the $3$ to divide the right side: $y=\dfrac{2}{3}x-\dfrac{7}{3}$ The slope-intercept form of the equation of a line is $y=mx+b$, where $m$ is its slope and $b$ is its $y$-intercept. For the equation given, $m=\dfrac{2}{3}$ Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{2}{3}$ The slope of the line whose equation must be found and a point through which it passes are now known. Substitute these elements into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ and simplify: $y-2=\dfrac{2}{3}[x-(-2)]$ $y-2=\dfrac{2}{3}(x+2)$ This is the equation of the line in point-slope form. Take all terms to the right side: $0=\dfrac{2}{3}(x+2)-y+2$ Simplify: $0=\dfrac{2}{3}x+\dfrac{4}{3}-y+2$ $0=\dfrac{2}{3}x-y+\dfrac{10}{3}$ Multiply the whole equation by $3$ and rearrange: $3\Big(0=\dfrac{2}{3}x-y+\dfrac{10}{3}\Big)$ $0=2x-3y+10$ $2x-3y+10=0$ This is the equation of the line in general form.