## Precalculus (6th Edition) Blitzer

For the point $\left( -3,-1 \right)$, the equation is $y+1=1\left( x+3 \right)$ and for the point $\left( 2,4 \right)$, the equation is $y-4=1\left( x-2 \right)$. The slope-intercept form of the line is $y=x+2$.
We have $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ as the points where the line passes. Here, these points are $\left( -3,-1 \right)$ and $\left( 2,4 \right)$. Hence, the slope of the line will be $m=\frac{4-\left( -1 \right)}{2-\left( -3 \right)}=\frac{5}{5}=1$. Thus, the slope of the line will be $1.$ The equation of the line in point-slope form will be calculated using the form $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ ………………………………………..….. (2) As there are two points through which the line passes, so there will be two equations of the line in point-slope form using equation (2). For the point $\left( -3,-1 \right)$, the equation will be $y+1=1\left( x+3 \right)$ ………………………………………..…… (3) For the point $\left( 2,4 \right)$, the equation will be $y-4=1\left( x-2 \right)$ ………………………………………..……. (4) The slope-intercept form can be concluded by simplifying either equation (3) or (4); we will get \begin{align} & y+1=1\left( x+3 \right) \\ & y+1-1=x+3-1 \\ & y=x+2 \end{align} Thus, the slope-intercept form of the line will be $y=x+2$.