Step-by-step explanation:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
where the slope m is:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
We have the points (2, -3) and (-6, -1).
Substitute:
[tex]m=\dfrac{-1-(-3)}{-6-2}=\dfrac{2}{-8}=-\dfrac{1}{4}[/tex]
[tex]y-(-3)=-\dfrac{1}{4}(x-2)\\\\y+3=-\dfrac{1}{4}(x-2)\to\text{point-slope form}[/tex]
The slope-intercept form of an equation of a line:
[tex]y=mx+b[/tex]
[tex]y+3=-\dfrac{1}{4}(x-2)[/tex]
[tex]y+3=-\dfrac{1}{4}x+\dfrac{1}{2}[/tex] subtract 3 from both sides
[tex]y=-\dfrac{1}{4}x-2\dfrac{1}{2}\to\text{slope-intercept form}[/tex]
The standard form of an equation of a line:
[tex]Ax+By=C[/tex]
[tex]y+3=-\dfrac{1}{4}(x-2)[/tex] multiply both sides by 4
[tex]4y+12=-(x-2)[/tex]
[tex]4y+12=-x+2[/tex] subtract 12 from both sides
[tex]4y=-x-10[/tex] add x to both sides
[tex]x+4y=-10\to\text{standard form}[/tex]
