Chapter 4.2, Problem 47HP

To determine

To Check:

Whether Tess and Jacinta have written the equation of line passing through points (3,-2) , (6,4) correct or not.

Answer to Problem 47HP

Jacinta is correct.

Tess is wrong.

Explanation of Solution

Given Information:

  $\begin{array}{l}Tess\\ m=\frac{4-(-2)}{6-3}=\frac{6}{3}=2\\ y=mx+b\\ 6=2(4)+b\\ 6=8+b\\ -2=b\\ y=2x-2\end{array}$

  $\begin{array}{l}Jac\mathrm{int}a\\ m=\frac{4-(-2)}{6-3}=\frac{6}{3}=2\\ y=mx+b\\ -2=2(3)+b\\ -2=6+b\\ -8=b\\ y=2x-8\end{array}$

Concept Used:

Equation of slope-intercept form of a line passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by:

  $y=(\frac{y_2-y_1}{x_2-x_1})x+b\mathrm{............................}(1)$

Where

b = y-intercept

Calculation:

First , we form the equation of line passing through points (3,-2) , (6,4):

To find slope of the line containing the given points

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\mathrm{.......................}[(x_1,y_1)=(3,-2),(x_2,y_2)=(6,4)]\\ m=\frac{4-(-2)}{6-3}=\frac{6}{3}=2\end{array}$

Use the slope and either of the two points to find the y-intercept:

Since the point lie on the line , they must satisfy the equation of line (1)

Let (x,y) = (3,-2)

  $\begin{array}{l}y=(\frac{y_2-y_1}{x_2-x_1})x+b\mathrm{............................}(1)\\ y=2x+b\mathrm{........................................}[m=2]\\ -2=2(3)+b\mathrm{...................................}[(x,y)=(3,-2)]\\ -2=6+b\mathrm{........................................}[simplify]\\ -2-6=6+b-\mathrm{6..............................}[subtract"6"each\_side]\\ -8=b\\ or\\ b=-8\end{array}$

Write the equation in slope intercept form:

  $\begin{array}{l}y=mx+b\\ y=2x-\mathrm{8.........................}[m=2,b=-8]\end{array}$

Hence , required equation is $y=2x-8$

So, answer of Jacinta is correct and Tess is wrong.

Reason:

Both of them calculated the slope correctly.

But Tess interchanged the x- and y-coordinates when she entered step 3 of solution.