Answer:
point slope form: y-3=[tex]-\frac{3}5}[/tex](x+2)
slope-intercept form: y=[tex]-\frac{3}{5}[/tex]x+[tex]\frac{9}{5}[/tex]
Step-by-step explanation:
so to find point-slope form, you first have to find the slope of the line
--slope =[tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex]=[tex]\frac{0-3}{3+2} =-\frac{3}{5}[/tex] (you have to use the two points shown: (-2,3) and (3,0) to plug in for y1, y2, x1, and x2 to find the slope.)
so your slope is [tex]-\frac{3}{5}[/tex].
Now all you have to do is use one of the points (-2,3) to sub in for y1 and x1 in the point-slope form equation.
Therefore, you get y-3=[tex]-\frac{3}{5}[/tex](x+2) for point-slope form.
Now for slope-intercept form, you already have the slope because you found it to complete the equation for point slope form
slope = [tex]-\frac{3}{5}[/tex]
Now slope intercept form is y=mx+b, where m is the slope and b is the y-intercept.
So, far we have y=[tex]-\frac{3}{5}[/tex]x+b
the easiet point to use for the x and y values is (3,0)
so, lets substitute: 0=[tex]-\frac{3}{5}[/tex](3)+b, which becomes 0=[tex]-\frac{9}{5}[/tex]+b
add [tex]-\frac{9}{5}[/tex] to both sides to get b=[tex]\frac{9}{5}[/tex]
Now you have both the slope and the y-intercept so you can now write the line in slope-intercept form.
You get [tex]y=-\frac{3}{5}x +\frac{9}{5}[/tex].
