# Math Help - Coordinate Geometry

1. ## Coordinate Geometry

Three points have coordinates A(-2,-1), B(10,6) and C(a,-6). Given AB=BC find the possible values of a. If the figure ABCD is a rhoMbus, find the coordinates of D.

2. Originally Posted by Ilsa
Three points have coordinates A(-2,-1), B(10,6) and C(a,-6). Given AB=BC find the possible values of a. If the figure ABCD is a rhoMbus, find the coordinates of D.
$AC= \sqrt{(-2-a)^2+ (-1+ 6)^2}= \sqrt{4+ 4a+ a^2+ 26}= \sqrt{a^2+ 4a+ 30}$
$BC= \sqr{(10- a)^2+ (6+ 6)^2}= \sqrt{100- 20a+ a^2+ 144}= \sqrt{a^2- 20a+ 244}$
Since AC= BC, we must have $a^2+ 4a+ 30}= a^2- 20a+ 244$. Subtracting $a^2- 20a+ 30$ from both sides leaves $24a= 214$.

The vector BC= <a, -6>- <10, 6>= <a- 10-, -12> is parallel to the vector AD. Find the coordinates of D by add BC to the corodinates of A.