Thanks EP. I will just add my 2c worth
The line appears to go through (1,0) and (-1,-1)
\(gradient=\frac{rise}{run}\\ gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\ gradient=\frac{y_2-y_1}{x_2-x_1}\\ gradient=\frac{-1-0}{-1-1}\\ gradient=\frac{-1}{-2}\\ gradient=\frac{1}{2}\\\)
now to get the equation I usually just think to myself
gradient =gradient
\(gradient=\frac{y_2-y_1}{x_2-x_1}\\ \text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\ \frac{1}{2}=\frac{y-0}{x-1}\\ 1(x-1)=2(y-0)\\ x-1=2y\\ x-2y-1=0 \)
(0,0) is in the wanted part of the graph so you have to sub in (0,0) and choose the correct inequality
(it is a dashed line, not a solid one, so there will be no equal sign)
\(LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\ LHS=0-0-1\\ LHS=-1\\ LHS<0\\ so\\ x-2y-1<0\)
LaTex:
gradient=\frac{rise}{run}\\
gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\
gradient=\frac{y_2-y_1}{x_2-x_1}\\
gradient=\frac{-1-0}{-1-1}\\
gradient=\frac{-1}{-2}\\
gradient=\frac{1}{2}\\
gradient=\frac{y_2-y_1}{x_2-x_1}\\
\text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\
\frac{1}{2}=\frac{y-0}{x-1}\\
1(x-1)=2(y-0)\\
x-1=2y\\
x-2y-1=0
LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\
LHS=0-0-1\\
LHS=-1\\
LHS<0\\
so\\
x-2y-1<0