Question 3452
To find the equation of a line that includes the points (6,6) and (2,4), use the equation for a line, {{{y = mx + b}}}, where m is the slope of the line and b is the intercept. For the sake of discussion, let's call (6,6) P1 and we'll call (2,4) P2<br>
First, find the slope. The slope is the "rise over the run" or the change in y divided by the change in x. We have two points on the line, so we know that the line "rises" from 4 to 6 (the y values of points P2 and P3). The point also "runs" from 2 to 6 (the x values of points P2 and P1). The distance of the rise is 6 - 4 or 2 and the distance of the run is 6 - 2 or 4. Therefore the slope is {{{m = 2/4 = 1/2}}}.<br>
With the slope, we have the equation  {{{y = (1/2)x + b}}}. Let's see what happens to the equation when we supply it with the x and y values on one point on the line, say P1. Our equation becomes {{{6 = (1/2)6 + b = 3 + b}}}. To find b, subtract 3 from both sides to get b = 3.<br>
The equation of the line is {{{y = (1/2) x + 3}}}.
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To verify this, put the x and y values of points P1 and P2 into the equation. If these points are on the line, the equation will hold.<br>
{{{ 6 = (1/2)6 + 3 = 3 + 3 = 6 }}} P1 holds<br>
{{{ 4 = (1/2)2 + 3 = 1 + 3 = 4 }}} P2 holds<br>