Answer
For the point $\left( 4,0 \right)$, the equation is $y-0=\frac{1}{2}\left( x-4 \right)$ and for the point $\left( 0,-2 \right)$, the equation is $y+2=\frac{1}{2}\left( x-0 \right)$. The slope intercept form of the line is $y=\frac{1}{2}x-2$.
Work Step by Step
The points through which the line passes are given by $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$. Here, these points are $\left( 4,0 \right)$ and $\left( 0,-2 \right)$.
Therefore, the slope of the line is
$m=\frac{-2-0}{0-\left( 4 \right)}=\frac{-2}{-4}=\frac{1}{2}$.
The slope of the line is $\frac{1}{2}.$
The equation of the line in point-slope form can be calculated by using the formula
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ …………………………………………….….. (2)
The line passes through two points, so there can be two equations of the line in point- slope form using equation (2).
For the point $\left( 4,0 \right)$ , the equation is
$y-0=\frac{1}{2}\left( x-4 \right)$ ………………………………………………… (3)
For the point $\left( 0,-2 \right)$ , the equation is
$y+2=\frac{1}{2}\left( x-0 \right)$ ………………………………………..……. (4)
The slope-intercept form can be given by simplifying either equation (3) or (4); we get
$\begin{align}
& y-0=\frac{1}{2}\left( x-4 \right) \\
& y=\frac{1}{2}x-2 \\
\end{align}$
The slope intercept form of the line is $y=\frac{1}{2}x-2$.
