1. ## 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!

2. ## 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.

3. ## 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 $\frac{22}{3}$.