Answer:
[tex]\huge\boxed{y=-\frac{5}{6}x-\frac{1}{3}}[/tex]
Useful Information:
The equation of a straight line: [tex]y=mx+c[/tex]
Step-by-step explanation:
To work this out you would first need to substitute the gradient into the equation, this gives you[tex]y=-\frac{5}{6}x+c[/tex].
The next step is to substitute the x and y coordinates from the point (2,-2) into the equation, this gives you [tex]-2=-\frac{5}{6}(2)+c[/tex]
In order to work out the value of c, you would have to bring the value of [tex]-\frac{5}{6}(2)[/tex] over to the other side, this can be done by adding [tex]\frac{5}{6}(2)[/tex] or [tex]-\frac{5}{3}[/tex] to -2, which gives you [tex]-\frac{1}{3}[/tex].
The final step is to substitute the m value of [tex]-\frac{5}{6}[/tex] and the c value of [tex]-\frac{1}{3}[/tex] into the equation, this gives you [tex]y=-\frac{5}{6}x-\frac{1}{3}[/tex]
1) Substitute the gradient.
[tex]y=-\frac{5}{6}x+c[/tex]
2) Substitute the x and y coordinates.
[tex]-2=-\frac{5}{6}(2)+c[/tex]
3) Bring [tex]-\frac{5}{6}(2)[/tex] over to the other side.
[tex]c=-2+\frac{5}{6}(2)[/tex]
4) Simplify to find the value of c.
[tex]c=-\frac{1}{3}[/tex]
5) Substitute the m and c values.
[tex]y=-\frac{5}{6}x-\frac{1}{3}[/tex]
