Answer
The equation of the line passing through $\left( 1,\ 6 \right)$ and $\left( -1,\ 2 \right)$ in point-slope and slope-intercept form is $y-2x-4=0$ and $y=2x+4$ respectively.
Work Step by Step
Two points on the line are given as $\left( 1,\ 6 \right)$ and $\left( -1,\ 2 \right)$.
Point-Slope form:
It is known that the equation of the line by point-slope form will be:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ ,
Where
$\begin{align}
& m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
& m=\frac{2-6}{-1-1} \\
& m=\frac{-4}{-2} \\
& m=2.
\end{align}$
Now, substituting the given data in the above equation to get the desired equation of the line:
$\begin{align}
& y-2=2\left\{ x-\left( -1 \right) \right\} \\
& y-2=2\left( x+1 \right) \\
& y-2=2x+2 \\
& y-2x-4=0.
\end{align}$
Slope-Intercept form:
The above equation can be rewritten as:
$y=2x+4$.
Compare the above equation with the general equation of slope-intercept form:
$\begin{align}
& y=mx+c \\
& \text{Therefore,} \\
& m=2 \\
& c=4.
\end{align}$
Therefore, the equation of the line passing through $\left( 1,\ 6 \right)$ and $\left( -1,\ 2 \right)$ in point-slope and slope-intercept form is $y-2x-4=0$ and $y=2x+4$ respectively.
