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Math Help - Coordinate Geometry

  1. #1
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    Coordinate Geometry

    Find the midpoint of the two intersection points of the graphs 2x-3y=7 and (x-6y)^2+9(3y-2)=-5.

    I could not understand the following method...

    Let the two intersection points be (X1, Y1) and (X2, Y2).

    Then X1+X2=22/3 and they work out Y1+Y2 using the value of X1+X2 and finally divide by 2 to get the midpoint.

    My question is how to get X1+X2=22/3?

    Your help will be much appreciated!
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  2. #2
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    Re: Coordinate Geometry

    Hi,
    The attachment shows why the technique works, but I don't see a way to immediately find the sum of the x values. Also, notice the caveat at the end.

    Coordinate Geometry-mhfintersection.png
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  3. #3
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    Re: Coordinate Geometry

    Well, you can get X1+ X2 by first getting X1 and X2! Those are the x coordinates of the points where 2x- 3y= 7 and (x- 6y)^2+ 9(3y- 2)= -5. Since it is x you ask about, solve the first equation for 3y= 2x- 7 so 6y= 4x- 14. Then (x- 6y)^2= (-3x+ 14)^2= 9x^2- 84x+ 196 and 9(3y- 2)= 18x- 63. So (x- 6y)^2+ 9(3y- 2)= 9x^2- 84x+ 196+ 18x- 63= 9x^2- 66x+ 133= -5 which then gives the quadratic equation 9x^2- 66x+ 138= 0 or 3x^2- 22x+ 46= 0. Solve that the two values of x or, more simply, recall that if X_1 and X_2 are the two solutions of ax^2+ bx+ c= 0 then X_1+ X_2= -\frac{b}{a} which, here, is \frac{22}{3}.
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