find the equation of the line containing the given points in slope-intercept form. Then, use the technique of this section to find a vector parametrization for the same line. Finally, show that your equations are equivalent.

$P(0,5),Q(2,-1)$

$P(0,5),Q(2,-1)$

The goal is to determine the line equation in both slope-intercept and vector parametrization form and prove that they are the same.

The slope-intercept form of the equation is, $y=mx+b$where $m$is the slope and $b$is the $y$intercept.

Find the slope of the given points first.

According to the formula, $m=\frac{y_2-y_1}{x_2-x_1}$

For the points, $P(0,5),Q(2,-1)$the slope is as follows,

$$\begin{gathered} m=\frac{-1-5}{2-0}\left[\text{since}{x}_1=0,y_1=5,x_2=2,y_2=-1\right] \\ m=\frac{-6}{2} \end{gathered}$$

$m=-3$

Now substitute the slope $m=-3$and the point $P(0,5)$in $y=mx+b$

$$\begin{gathered} 5=-3·0+b \\ 5=b \\ b=5 \end{gathered}$$

By substitution of $m=-3,b=5$in the equation $y=mx+b$

$y=-3·x+5$

Thus the line equation is $y=-3x+5$

Now find the line equation using the vector function.

The points are $P(0,5),Q(2,-1)$

We'll start by determining the direction vector of the line $\overrightarrow{PQ}$

The points are $P(0,5)$and $Q(2,-1)$

$$\begin{gathered} \overrightarrow{PQ}=(2-0,-1-5) \\ \overrightarrow{PQ}=(2,-6) \end{gathered}$$