Question 207652: hello tutors. i need your help with this problem.
find an equation of the line that passes through the origin and the point (4,-8)
thank you so much, my memory is very rusty and i could really use a refresher!
Found 2 solutions by rfer, Theo:
Answer by rfer(16322)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! equation of straight line is y = mx + b where m is the slope and b is the y intercept.
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if the line passes through the origin, then 1 point on the line is (0,0) and the other point is (4,-8) as given.
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slope is (y2-y1)/(x2-x1)
if we let (x1,y1) = (0,0) and we let (x2,y2) = (4,-8), then the slope equation becomes
(-8-0)/(4-0) = (-8)/(4) = -2.
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to solve for the y intercept we can use either point and substitute it in the equation of y = -2*x + b
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assuming (0,0) is the point, that equation becomes 0 = -2*0 + b making b = 0
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assuming (4,-8) is the point, that equation becomes -8 = -2*4 + b which becomes -8 = -8 * b which becomes b = 0 once we add 8 to both sides of the equation.
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the equation of your line should be y = 2*x
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a graph of your equation will look like this.
when x is 0, y is 0.
when x is 1, y is -2.
when x is 2, y is -4.
when x is -1, y is 2.
when x is 2, y is 4.
etc., as shown on the graph.
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