Question 1118579
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The matrix/determinant solution given by tutor @alan3354 is useful for more complex problems, but it seems a lot of work for this rather elementary problem.<br>
Both of the other tutors who responded used a formula to find the slope of the line and then used the point-slope form of the equation to find the specific equation of the line in your example.<br>
Those are valid ways to solve the problem.  But I have some suggestions for alternative methods that you might want to try.<br>
First:  Finding the slope determined by the two given points.<br>
There is a simple formula for slope: {{{m = (y2-y1)/(x2-x1)}}}.<br>
Yes, a simple formula; but it is easy to get numbers in the wrong places and end up with the wrong slope, which of course makes any subsequent work you do on the problem a waste of time.<br>
I always suggest to students that they at least mentally, if not on paper, make a sketch of the two points and use the sketch to find the slope by comparing the x and y coordinates of the two given points.  Specifically, for the given points (-1,-13) and (1,3) in your problem, I think the safest way to find the slope is as follows:<br>
Slope is the ratio of how fast the graph changes vertically and how fast it changes horizontally: "slope = rise / run".  More specifically, I always think of it as measuring how far the graph goes up or down each time I move 1 unit to the right.  So<br>
(1) Since I always think of moving left to right, my "first" point will be the one that is farther left -- i.e., has the smaller x coordinate.  So (-1,-13) is my "first" point and (1,3) is my second point.<br>
(2) From the first point to the second, the x value changes from -1 to +1, a change of 2: the graph moved 2 units horizontally; the "run" is 2.<br>
(3) From the first point to the second, the y value changes from -13 to +3, a change of 16: the graph moved up 16 units; the "rise" is 16.<br>
(4) The slope is "rise"/"run" = 16/2 = 8.<br>
You will make far fewer mistakes in finding slopes of lines if you use this method instead of plugging numbers into a magic formula.<br>
Second: Finding the equation of the line<br>
While use of the point-slope form of the equation of a line to finish finding the specific equation for a particular example is fine, in my experience more students find it easier to use the slope-intercept form.  For your particular problem, it would go like this:<br>
Plug the given x and y coordinates of one of the given points into the basic slope-intercept equation and solve for b:<br>
{{{y = mx+b}}}
{{{3 = 8(1)+b}}}
{{{3 = 8+b}}}
{{{b = -5}}}<br>
Now you have both the slope and the y-intercept; the equation of the line is y = 8x-5.