Question

Find the equation (in terms of x) of the line through the points (-4,3) and (3,1)y =

Answer 1

The Slope-Intercept Form of the equation of a line is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

So, given the points:

$\begin{gathered} \mleft(-4,3\mright) \\ \mleft(3,1\mright) \end{gathered}$

You can find the slope by using this formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

In this case, you can set up that.

$\begin{gathered} y_2=3 \\ y_1=1 \\ \\ x_2=-4 \\ x_1=3 \end{gathered}$

Then, substituting values into the formula and evaluating, you get:

$m=\frac{3-1}{-4-3}=\frac{2}{-7}=-\frac{2}{7}$

Now you can substitute the slope of the line and the coordinate of one of the points given in the exercise, into the equation

$y=mx+b$

Then, substituting the coordinates of the point:

$(3,1)$

And then solving for "b", you get that this is:

$\begin{gathered} 1=(-\frac{2}{7})(3)+b \\ \\ 1=-\frac{6}{7}+b \\ \\ 1+\frac{6}{7}=b \\ \\ b=\frac{13}{7} \end{gathered}$

Finally, knowing "m" and "b", you can determine that the equation of this line (in terms of "x"), is:

$\begin{gathered} y=mx+b \\ \\ y=-\frac{2}{7}x+\frac{13}{7} \end{gathered}$

The answer is:

$y=-\frac{2}{7}x+\frac{13}{7}$