Math 150 – Fall 2015 Section 3A 1 of 6 Section 3A – Rectangular Coordinate System Rectangular (or Cartesian) Coordinate System: • We refer to this as the plane (or 2-dimensional plane) • The origin is the intersection of the x-axis and y-axis. • Each point in the plane corresponds with an ordered pair of real numbers (x, y). • To plot a point, travel horizontally from the origin the distance and direction given by the x-coordinate (positive is right, negative is left), then vertically from the x − axis the distance and direction given by the y-coordinate (positive is up, negative is down). Quadrants: The x-axis and y−axis divide the plane into four quadrants. The first quadrant is where x and y are both positive (upper right), and then they are numbered counter-clockwise. We always use Roman Numerals I, II, III, and IV to number the quadrants. Example 1. Plot the following points on a rectangular coordinate system: A(3,4) B(-3,-5) C(9,-9) D(0,-3) E(-5,0) Example 2. Shade the region of the coordinate plane that contains the set of ordered pairs {(x, y)|y < 1} (read this as “the set of all ordered pairs (x, y) such that y > 1”) Math 150 – Fall 2015 Section 3A 2 of 6 Example 3. Shade the region of the plane that contains the ordered pairs {(x, y)| − 4 ≤ x ≤ 2, y > −2} Definition. • Any set of ordered pairs is called a relation. • The plot of every point associated with an ordered pair is called the graph of the relation. • The set of all first elements in the ordered pairs is called the domain of the relation. • The set of all second elements in the ordered pairs is called the range of the relation. Example 4. Refer back to Example 1, where we plotted five points. If we consider this as a relation, then • Write the relation as a set: • Give the domain written as a set: • Give the range written as a set: • The graph would be the graph we did earlier. Example 5. For the relations in Examples 2 and 3, find the domain and range. Example 2: • Domain: • Range: Example 3: • Domain: • Range: Math 150 – Fall 2015 Section 3A 3 of 6 Distance Formula Pythagorean Theorem. For a right triangle with legs of length a and b and hypotenuse of length c, then a2 + b2 = c2 Pythagorean Theorem Graph Distance Formula Graph Theorem. The distance between two points P (x1 , y1 ) and Q(x2 , y2 ), called d(P, Q), is given by p d(P, Q) = (x1 − x2 )2 + (y1 − y2 )2 Example 6. Find the distance between the points A(2, 5) and B(−3, 9). Example 7. Determine whether the points A(5, 3) and B(−1, −1) are equidistant from C(2, 1). Example 8. If the distance between A(8, x) and B(4, 7) is 5 units, find all possible values for x. Example 9. Determine whether the points A(1, −3), B(−2, 0), and C(2, −1) are the vertices of a right triangle. Math 150 – Fall 2015 Section 3A 4 of 6 Midpoint Formula Definition. The midpoint, M(x,y), of a line segment is the point that lies halfway between the two endpoints on the line segment that connects them. Theorem. The midpoint of the line segment that connects A(x1 , y1 ) and B(x2 , y2 ) is the point M (x, y) with coordinates x1 + x2 y1 + y2 M (x, y) = , 2 2 Example 10. Find the midpoint of the line segment that connects the points (−3, 8) and (6, −4). Example 11. If (6, −3) is the endpoint of the line segment connecting A(2, 3) and B, then find the coordinates of the other endpoint B. Circles Definition. The set of all points P (x, y) that are a given fixed distance from a given fixed point is called a circle. • The fixed distance, r, is called the radius. • The fixed point C(h, k) is called the center. Theorem. The equation of a circle with center C(h, k) and radius r is (x − h)2 + (y − k)2 = r2 . This is called the standard form. If the center of the circle is the origin C(0, 0) then we have the formula x2 + y 2 = r 2 . Math 150 – Fall 2015 Section 3A 5 of 6 Example 12. Write the equation of the circle with center C(−4, 3) and radius r = 5 in standard form. Graph the circle and give its domain and range. Example 13. Find the center and radius of the circles. Also, give each circle’s domain and range. (a) x2 + (y − 3)2 = 36 (b) (x + 2)2 + (y + 7)2 − 13 = 9 General Form of an Equation of a Circle: The general form of an equation of a circle is x2 + y 2 + cx + dy + e = 0 This is always a circle, but it is not easy to identify the center and radius like with the standard form. Changing to Standard Form: If we have an equation of a circle in general form, then we can rewrite it in standard form by: 1. Group the x and y terms together. 2. Move the constant to the right-hand side of the equation. 3. Complete the square for the x terms. 4. Complete the square for the y terms. Example 14. Find the center and radius of the following circles. (a) x2 + y 2 − 3x + 11x − 4 = 0 Math 150 – Fall 2015 Section 3A 6 of 6 (b) 5x2 + 5y 2 + 7x − 8y − 17 = 0 Example 15. Write an equation that describes all the points (x, y) that are 7 units away from point B(4, −6). Example 16. Suppose the points A(−3, 3) and B(5, 7) are endpoints of a diameter of a circle. Find the center of the circle and the radius. Write the equation of the circle in standard form.