Respuesta :
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad
(\stackrel{x_2}{1}~,~\stackrel{y_2}{2})
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-1}{1-0}\implies \cfrac{1}{1}\implies 1
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-1=1(x-0)\implies y-1=x
\\\\\\
\stackrel{standard~form}{-x+y=1}[/tex]
Hello!
First we find the slope, which is the the difference of the y-values divided by the difference of the x-values as seen below.
[tex] \frac{2-1}{1-0} = \frac{1}{1}= 1[/tex]
The slope of our line is 1.
Now, the y-intercept is where x=0 is located at. As you can see, we already have a point with x=0. Therefore our y-intercept is 1, as the y-value is in that ordered pair.
Just to check, we will plug in a point from our line (1,2) in the slope intercept equation and solve for b.
2=1(1)+b
2=1+b
b=1
We can write our equation in standard form (Ax+By=C) below.
-x+y=1
I hope this helps!
First we find the slope, which is the the difference of the y-values divided by the difference of the x-values as seen below.
[tex] \frac{2-1}{1-0} = \frac{1}{1}= 1[/tex]
The slope of our line is 1.
Now, the y-intercept is where x=0 is located at. As you can see, we already have a point with x=0. Therefore our y-intercept is 1, as the y-value is in that ordered pair.
Just to check, we will plug in a point from our line (1,2) in the slope intercept equation and solve for b.
2=1(1)+b
2=1+b
b=1
We can write our equation in standard form (Ax+By=C) below.
-x+y=1
I hope this helps!
