Solving Eq & Ineq        Graphs & Func.        Systems of Eq.        Polynomials        Frac. Express.        Powers & Roots        Complex Numbers        Quadratic Eq.        Quadratic Func.   Coord. Geo.        Exp. & Log. Func.        Probability        Matrices        Trigonometry        Trig. Identities        Equations & Tri. On this page, we hope to clear up any problems you might have with coordinate geometry (circles, ellipses, midpoints, etc.).  If you're in to graphs, this is your section! Distance and midpoint formulas Circles Ellipses Hyperbolas Systems of equations Quiz on Coordinate Geometry When dealing with lines and points, it is very important to be able to find out how long a line segment is or to find a midpoint.  However, since the midpoint and distance formulas are covered in most geometry courses, you can click here to better your understanding of the midpoint and distance formulas. Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin.  When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2.  Example: 1. Problem: Find the center and radius of (x - 2)2 + (y + 3)2 = 16. Then graph the circle. Solution: Rewrite the equation in standard form. (x - 2)2 + [y - (-3)]2 = 42 The center is (2, -3) and the radius is 4. The graph is easy to draw, especially if you use a compass. The figure below is the graph of the solution. Ellipses, or ovals, when centered at the origin, have an equation (standard form) of (x2/a2) + (y2/b2) = 1.  When the center of the ellipse is at (h, k), the equation (in standard form) is as follows: (x - h)2 (y - k)2 -------- + -------- = 1 a2 b2 Example: 1. Problem: Graph x2 + 16y2 = 16. Solution: Multiply both sides by 1/16 to put the equation in standard form. x2 y2 -- + -- = 1 16 1 a = 4 and b = 1. The vertices are at (±4, 0) and (0, ±1). (The points are on the axes because the equation tells us the center is at the origin, so the vertices have to be on the axes.) Connect the vertices to form an oval, and you are done! The figure below is the graph of the ellipse. The equation of a hyperbola (in standard form) centered at the origin is as follows: x2 y2 -- - -- = 1 a2 b2 Example: 1. Problem: Graph 9x2 - 16y2 = 144. Solution: First, multiply each side of the equation by 1/144 to put it in standard form. x2 y2 -- - -- = 1 16 9 We now know that a = 4 and b = 3. The vertices are at (±4, 0). (Since we know the center is at the origin, we know the vertices are on the x axis.) The easiest way to graph a hyperbola is to draw a rectangle using the vertices and b, which is on the y-axis. Draw the asymptotes through opposite corners of the rectangle. Then draw the hyperbola. The figure below is the graph of 9x2 - 16y2 = 144. The easiest way to solve systems of equations that include circles, ellipses, or hyperbolas, is graphically.  Because of the shapes (circles, ellipses, etc.), there can be more than one solution.  Example: 1. Problem: Solve the following system of equations: x2 + y2 = 25 3x - 4y = 0 Solution: Graph both equations on the same coordinate plane. The points of intersection have to satisfy both equations, so be sure to check the solutions. Both intersections do check. The figure below shows the solution. Take the Quiz on coordinate geometry.  (Very useful to review or to see if you've really got this topic down.)  Do it!

Math for Morons Like Us - Algebra II: Coordinate Geometry
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