Answer
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 $y=x+2$.
Work Step by Step
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 $y=x+2$.
