## College Algebra (6th Edition)

Point-slope form: $y-3=\dfrac{3}{2}(x+1)$ General form: $3x-2y+9=0$
Passing through $(-1,3)$ and parallel to the line whose equation is $3x-2y-5=0$ Rewrite the equation given in slope-intercept form by solving it for $y$: $3x-2y-5=0$ $3x-5=2y$ $2y=3x-5$ $y=\dfrac{3}{2}x-\dfrac{5}{2}$ Since the slope-intercept form of the equation of a line is $y=mx+b$, the slope of the given equation can be identified as $m=\dfrac{3}{2}$ Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{3}{2}$ The slope of the line whose equation must be found and a point through which it passes are now known. Substitute them into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ $y-3=\dfrac{3}{2}[x-(-1)]$ $y-3=\dfrac{3}{2}(x+1)$ This is the point-slope form of the equation of the line. Take all terms to the right side and simplify: $0=\dfrac{3}{2}(x+1)-y+3$ $0=\dfrac{3}{2}x+\dfrac{3}{2}-y+3$ $0=\dfrac{3}{2}x-y+\dfrac{9}{2}$ Rearrange: $\dfrac{3}{2}x-y+\dfrac{9}{2}=0$ Multiply the whole equation by $2$: $2\Big(\dfrac{3}{2}x-y+\dfrac{9}{2}=0\Big)$ $3x-2y+9=0$ This is the general form of the equation of the line