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Question 815705: The abscissa of a point is two-fifths of its ordinate. Find the point if it is 5 units from (-2,2).
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! point1 (x,y)
x = (2/5)y
y = (5/2)x
point2 (-2,2)
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distance between two points:
d = sqrt( dx^2 + dy^2 )
where:
dx is the difference between the x-coordinates
dy is the difference between the y-coordinates
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d = sqrt( dx^2 + dy^2 )
5 = sqrt( (x + 2)^2 + ((5/2)x - 2)^2 )
25 = (x + 2)^2 + ((5/2)x - 2)^2
25 = (x + 2)(x + 2) + ((5/2)x - 2)((5/2)x - 2)
25 = xx + 4x + 4 + (25/4)xx - 10x + 4
25 = (4/4)xx + (25/4)xx - 6x + 8
(29/4)xx - 6x - 17 = 0
7.25xx - 6x - 17 = 0
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the above quadratic equation is in standard form, with a=7.25, b=-6, and c=-17
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to solve the quadratic equation, plug this:
7.25 -6 -17
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the roots of the quadratic are:
x = 2
x = -1.17241379
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Answer:
both roots are valid values of x for point1
(i.e. both roots answer the question as to what is the value of x for point1.)
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point1: (2, 5)
or
point1: (-1.17241379, -2.9310343)
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Solve quadratic equations, quadratic formula:
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