Find the coordinates of the mid-point of the straight line joining the points of intersection of the curve $\displaystyle x^2+2y^2+5x=68$ and the line $\displaystyle 2y+3x=9$

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- Aug 21st 2010, 02:52 AMPunchCoordinates of the midpoint of intersecting lines
Find the coordinates of the mid-point of the straight line joining the points of intersection of the curve $\displaystyle x^2+2y^2+5x=68$ and the line $\displaystyle 2y+3x=9$

- Aug 21st 2010, 03:14 AMProve It
$\displaystyle x^2 + 2y^2 + 5x = 68$ and $\displaystyle y = \frac{9-3x}{2}$.

Substituting the second into the first gives

$\displaystyle x^2 + \left(\frac{9-3x}{2}\right)^2 + 5x = 68$

$\displaystyle \frac{4x^2}{4} + \frac{81 -54x + 9x^2}{4} + \frac{20x}{4} = \frac{272}{4}$

$\displaystyle \frac{13x^2 - 34x - 191}{4} = 0$

$\displaystyle 13x^2 - 34x - 191 = 0$

Now solve for $\displaystyle x$ and substitute back to find the corresponding $\displaystyle y$ values. This will give you the two points, so you can use the midpoint rule to find the midpoint. - Aug 21st 2010, 03:14 AMHallsofIvy
The kind of obvious thing to do is to find the two point of intersection. From 2y+ 3x= 9, y= 9/2- (3/2)x. Replace y in $\displaystyle x^2+ 2y^2+ 5x= 68$ and you get a quadratic equation for x which should have two solutions. Once you have found those two points of intersection, find the point midway between them.

Do you know that the point midway between $\displaystyle (x_0, y_0)$ and $\displaystyle (x_1, y_1)$ is $\displaystyle \left(\frac{x_0+ x_1}{2}, \frac{y_0+ y_1}{2}\right)$? - Aug 21st 2010, 03:55 AMPunch
Sorry guys, I misunderstood that the question stated there was only 1 point of intersection...

didnt notice that the word was 'points'.