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Question 45318: find the equation of the line that passes through (-2,3) and (4,-1).
the formula I believe is y-y1 = y2-y1/x2-x1 (x-x1)
I'm not quite sure as to how to set that up. do I solve as I go?
Found 2 solutions by tutmath, tutorcecilia:
Answer by tutmath(14)   (Show Source):
You can put this solution on YOUR website! The formula is correct.
(x1, y1) = (-2, 3)
(x2, y2) = (4, -1)
Now substitute the above values in the formula:
(x-x1)/(x2-x1) = (y-y1)/(y2-y1)
we get:
(x-(-2))/(4-(-2)) = (y-3)/(-1-3)
that is:
x+2/6 = y-3/-4
-4(x+2) = 6(y-3)
-4x - 8 = 6y - 18
Thus the equation of the line is:
4x + 6y - 10 = 0
The way to verify this it to substitute the two values above in place of x and y and check that the equation is true.
Example for (-2,3): 4*(-2) + 6*3 - 18 = 0 which is true.
(If you have any questions on the solution you can send me your queries at tutmath@gmail.com)
Answer by tutorcecilia(2152)  (Show Source):
You can put this solution on YOUR website! First, I would find the slope of the line:
m = (y2-y1)/(x2-x1) = (-1-3)/4-(-2) = -4/6 = -2/3
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Use the slope-intercept formula (y=mx+b) to find the y-intercept(b).
Pick point (-2, 3) and plug-in the values:
(3)=(-2/3)(-2)+ b
3 = 4/3 + b
3 - 4/3 = 4/3 - 4/3 + b
(9-4)/3 = b
b = 5/3
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Plug the slope (m) and the y=intercept (b) into the formula y-mx+b:
y=(-2/3)x + 5/3
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Check the formula by plugging in one of the given points. Pick point (4, -1):
(-1)=(-2/3)(4)+ 5/3
-1=-3/3
-1=-1 Checks out.
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