SOLUTION: compute x such that a straight line passing thru (2,x)and x-2)has has a slope of -2/3

Question 173422: compute x such that a straight line passing thru (2,x)and x-2)has has a slope of -2/3
Found 2 solutions by Alan3354, gonzo:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
compute x such that a straight line passing trhu (2,x)and x-2)has has a slope of -2/3
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m = (y2-y1)/(x2-x1)
-2/3 = (-2-x)/(x-2)
-2 = (-6-3x)/(x-2)
-2x + 4 = -6 - 3x
x = -10

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
slope intercept form of equation is:
y = m*x + b
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m = slope
b = y intercept( value of y when x = 0)
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m = -2/3 (given)
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the equation to find the slope is:
slope (m) = (y2-y1)/(x2-x1)
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you have provided me with one of the x values (2)
you have also provided me with two of the y values:
those are: x and x-2
i will call them by a name other than x to avoid confusion.
instead of x and x-2, call them k and k-2.
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so we have two points:
(x1,y1) = (2,k)
(x2,y2) = (x2,k-2)
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we can plug these values into the formula for the slope:
(y2-y1) / (x2-x1) becomes:
[(k-2)-k] / (x2-2)
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since the slope = -2/3, then:
-2 = (k-2)-k)
and:
x2-2 = 3
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solving for k, we get:
-2 = k-2-k
-2 = -2
this is a true statement but we still don't know what k is since any value of k will satisfy this equation (k cancelled out).
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solving for x2, we get:
x2 - 2 = 3
x2 = 5
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our two points have now become:
(x1,y1) = (2,k)
(x2,y2) = 5,k-2)
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since k can be any value, you don't have enough to get one line.
you can get an infinite number of lines depending on what value you assign to k.
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in order to find the equation of a specific line, you need a minimum of two things:
the slope (you have that)
one point (you don't have that).
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unless i read your problem wrong, you don't have enough informtion to pinpoint the equation for one line.
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