SOLUTION: A line passes through (-2,3), (2,5), and (6,k). Find k.

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Question 69071This question is from textbook Algebra Stucture and Method Book 1
: A line passes through (-2,3), (2,5), and (6,k). Find k. This question is from textbook Algebra Stucture and Method Book 1

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start by finding the slope through the points (-2,3), (2,5). You can use the formula:
%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 which is simply a formal way to represent "rise" over "run". By plugging the points (-2,3), (2,5) into the formula it becomes
%285-3%29%2F%282-%28-2%29%29
which simplifies to
2%2F4=1%2F2
This number represents the slope of the line running through the points (-2,3), (2,5), and (6,k). To complete the equation of the line, we need to plug in one of those points. For instance, we can use the general form of the line equation: y=mx+b, where m is the slope and b is the y-intercept, and solve for b. So lets plug in the point (-2,3) in the equation along with the given slope:
y=mx%2Bb using (-2,3) and m=1/2
3=%281%2F2%29%28-2%29%2Bb solve for b
3=-1%2Bb isolate b
4=b
so the equation of the line is y=1/2x+4
Now all we need to do is plug in the point (6,k) to find k.
y=%281%2F2%29%286%29%2B4
y=3%2B4
y=7
so k=7 and the point is (6,7)