Q 38. - Precalculus Enhanced with Graphing Utilities

Question

Make up four problems that you might be asked to do given the two points $\left(-3,4\right)$ and $\left(6,1\right)$. Each problem should involve a   different concept.Be sure that the directions are clearly stated.

(a) Find the equation of line .Graph the line 

(b) Find the slope for the equation of line .

(c) Find the center between two points.

(d) Find the distance between given points.

Step-by-Step Solution

Answer

(a) Equation of line is $x+3y=15$

(b) Slope of the line is $m=\frac{-1}{3}$

(c) Midpoints of center are $\left(\frac{3}{2} ,\frac{5}{2}\right)$

(d) Distance between the points is $\sqrt{90}$.

Step 1. Given information.

Consider the given points $\left(-3,4\right)$ and $\left(6,1\right)$.

Step 2. (a) Formula used.

The formula used to find equation of line is $y-y_1=m\left(x-x_1\right)$ and to find slope $m=\frac{y_2-y_1}{x_2-x_1}$.

$m=\frac{1-4}{6+3}$

$m=\frac{-1}{3}$

Substitute the value in equation.

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-4=\frac{-1}{3}\left(x+3\right) \\ x+3y=15 \end{gathered}$$

Step 3. Plot the graph.

The graph for the equation $x+3y=15$ is shown below.

Step 4. (b) Formula used to find slope .

The formula to find slope is$m=\left(\frac{y_2-y_1}{x_2-x_1}\right)$.

Substitute the value of $x_1=-3 ,y_1=4$ and $x_2=6 ,y_2=1$.

$$\begin{gathered} m=\frac{1-4}{6+3} \\ m=\frac{-1}{3} \end{gathered}$$

Step 5. (c) Formula used to find midpoints.

The formula used to find midpoint is $\left(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}\right)$.

Substitute the values .

$$\begin{gathered} \left(\frac{-3+6}{2} ,\frac{4+1}{2}\right) \\ \left(\frac{3}{2},\frac{5}{2}\right) \end{gathered}$$

Step 6. (d) Formula used to find distance.

The formula used to find distance is$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Substitute the value of $x_{1 },y_1$ and $x_2 ,y_2$.

$$\begin{gathered} d=\sqrt{{\left(6+3\right)}^2+{\left(1-4\right)}^2} \\ d=\sqrt{90} \end{gathered}$$