Answer:
[tex]y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)[/tex]
OR
[tex]y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)[/tex]
Step-by-step explanation:
Hi there!
Point-slope form: [tex]y-y_1=m(x-x_1)[/tex] where [tex](x_1,y_1)[/tex] is a point and [tex]m[/tex] is the slope
1) Determine the slope
[tex]m=\frac{\displaystyle y_2-y_1}{\displaystyle x_2-x_2}[/tex] where two given points are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]
Plug in the given points (-2, 6) and (3,-2):
[tex]m=\frac{\displaystyle -2-6}{\displaystyle 3-(-2)}\\\\m=\frac{\displaystyle -8}{\displaystyle 3+2}\\\\m=-\frac{\displaystyle 8}{\displaystyle 5}[/tex]
Therefore, the slope of the line is [tex]-\frac{\displaystyle 8}{\displaystyle 5}[/tex]. Plug this into [tex]y-y_1=m(x-x_1)[/tex]:
[tex]y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)[/tex]
2) Plug in a point [tex](x_1,y_1)[/tex]
[tex]y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)[/tex]
We're given two points, (-2, 6) and (3,-2), so there are two ways we can write this equation:
[tex]y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x-(-2))\\\\y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)[/tex]
OR
[tex]y-(-2)=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)\\y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)[/tex]
I hope this helps!
