Q 22. - Calculus

Question

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. 

$P(3,-2), Q(6, 4)$

Step-by-Step Solution

Answer

Slope intercept form $y=2 x-8 ; x\left(t\right)=3 t+3, y\left(t\right)=-2+6 t$ equivalent

Step 1: Given information

$P(3,-2), Q(6,4)$

Step 2: Calculation

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

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

Here first find the slope of the given points.

According to the formula, $m=\frac{y2-y1}{x2-x1}$

For the points, $P(3,-2), Q(6,4)$ the slope is as follows,

$m=\frac{4-(-2)}{6-3}\left[\text{ since }{x}_1=3,y_1=-2,x_2=6,y_2=4\right]$

$$\begin{gathered} m=\frac{6}{3} \\ m=2 \end{gathered}$$

Now substitute the slope $m=2$ and the point $P(3,-2)$ in $y=m x+b$

$$\begin{gathered} -2=2·3+b \\ -2=6+b \end{gathered}$$

On both sides of the equation, add $-6$

$$\begin{gathered} -2-6=6+b-6 \\ b=-8 \end{gathered}$$

By substitution of $m=2, b=-8$ in the equation $y=m x+b$

$$\begin{gathered} y=2·x+(-8) \\ y=2x-8 \end{gathered}$$

Thus the line equation is $y=2 x-8$

Step 3: Calculation

Now, using the vector function, calculate the line equation.

The points are $P(3,-2), Q(6,4)$

First we will find the direction vector for the line $\overrightarrow{PQ}$

The points are $P(3,-2)$ and $Q(6,4)$

$$\begin{gathered} \overrightarrow{PQ}=(6-3,4-(-2\left)\right) \\ \overrightarrow{PQ}=(3,6) \end{gathered}$$

The formula to find the line $\mathcal{L}$ equation is as follows, $r\left(t\right)=P_0+td$ Where, $P_0$ is the point and $d$ is the direction vector.

Here $P(3,-2)$ and $\overrightarrow{PQ}=d=(3,6)$ then the equation is,

$$\begin{gathered} r\left(t\right)=(3,-2)+t(3,6) \\ r\left(t\right)=(3+3t,-2+6t) \end{gathered}$$

The equation is written as follows,

$r\left(t\right)=(3+3 t,-2+6 t)$

The equation of a line $\mathcal{L}$ in the form of vector parametrization is, $r\left(t\right)=(3+3 t,-2+6 t)$ $x\left(t\right)=3+3 t, y\left(t\right)=-2+6 t$

By eliminating the parameter $t$ from vector parametrization $y=2 x-8$ the equation becomes equals to the slope-intercept form of the equation $y=2 x-8$

The vector parametrization $r\left(t\right)=(3+3 t,-2+6 t)$

$x=3+3 t, y=-2+6 t$

Substitute $t=\frac{x-3}{3}$ in $y=-2+6t.\left[\right.$ since $\left.x=3+3t\Rightarrow 3t=x-3\Rightarrow x=\frac{x-3}{3}\right]$

Then, $y=-2+6·\frac{x-3}{3}$

$y=-2+2·(x-3)$

$y=2 x-8$ Which is equal to the slope-intercept form.

As a result, the equation's slope-intercept form and the vector parametrization equations are equal.

$y=2 x-8$ is the slope-intercept form, and $r\left(t\right)=(3+3 t,-2+6 t)$ is the vector parametrization.

Therefore, the required equation in slope-intercept form$y=2 x-8 ; x\left(t\right)=3 t+3, y\left(t\right)=-2+6 t$ equivalent .