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Distance and Midpoint Formulas; Circles

Distance and Midpoint Formulas; Circles. The Midpoint Formula. Consider a line segment whose endpoints are ( x 1 , y 1 ) and ( x 2 , y 2 ). The coordinates of the segment's midpoint are To find the midpoint, take the average of the two x -coordinates and of the two y -coordinates.

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Distance and Midpoint Formulas; Circles

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  1. Distance and Midpoint Formulas; Circles

  2. The Midpoint Formula • Consider a line segment whose endpoints are (x1, y1) and (x2, y2). The coordinates of the segment's midpoint are • To find the midpoint, take the average of the two x-coordinates and of the two y-coordinates.

  3. Example Find the midpoint of the line segment with endpoints (1, -6) and (-8, -4). To find the coordinates of the midpoint, we average the coordinates of the endpoints. (-7/2, -5) is midway between the points (1, -6) and (-8, -4). **Find the midpoint of the line segment with end points (5,2) and (7,8).

  4. Distance Formula The distance between two points on a number line is |a-b| or |b-a|. We can derive the Distance Formula by using this information and the Pythagorean Theorem.

  5. The Distance Formula • The distance, d, between the points (x1, y1) and (x2,y2) in the rectangular coordinate system is

  6. Example Find the distance between (-1, 2) and (4, -3). Letting (x1, y1) = (-1, 2) and (x2, y2) = (4, -3), we obtain

  7. Example What is the distance between P(-1,4) and Q(2,-3)?

  8. Center (h, k) Any point on the circle Radius: r (x, y) Circles Circle- is the set of all points in a plane that are equidistant from a given point in the plane called the center. Radius – any line segment whose endpoints are the center and any point on the circle.

  9. Circles Equation of a circle with center (h, k) and radius r is • Find the center and radius of • Write an equation given that the center is at (0,3) and the radius =7. • Write an equation given that the endpoints of a diameter are (2,8) and (2,-2):

  10. (x – 2)2 + (y – (-3))2 = 32 h is 2. k is –3. r is 3. Example Find the center and radius of the circle whose equation is (x – 2)2 + (y + 3)2 = 9 and graph the equation. In order to graph the circle, we need to know its center, (h, k), and its radius r. We can find the values of h, k, and r by comparing the given equation to the standard form of the equation of a circle. (x – 2)2 + (y + 3)2 = 9

  11. 3 2 1 1 2 3 4 5 -1 -2 (2, -3) -3 3 -4 -5 -6 Example cont. Find the center and radius of the circle whose equation is (x – 2)2 + (y + 3)2 = 9 and graph the equation. We see that h = 2, k = -3, and r = 3. Thus, the circle has center (2, -3) and a radius of 3 units. Plot the center, (2, -3), and find 3 additional points by going up, right, down, and left of the center by 3 units.

  12. Graph x2+ y2 = 25

  13. (x,y) (x,y) r r (h,k) (h,k) (h,k) (h,k) r r (x,y) (x,y) No matter where the circle is located, or where the center is, the form of the equation is the same.

  14. The radius is (x - -1)2 + (y - 4)2 = Example • Find the equation of the circle whose diameter has endpoints of (-5, 2) and (3, 6). Therefore the center is (-1, 4). (x + 1)2 + (y - 4)2 = 20

  15. Example Considerx2 + y2 + 6x - 2y - 54 = 0 . x2 + y2+ 6x - 2y = 54 x2 + 6x + y2- 2y = 54 (6/2) = 3 (-2/2) = -1 (3)2 = 9 (-1)2 = 1 x2 + 6x + 9 + y2 - 2y + 1 = 54 + 9 + 1 (x + 3)2 + (y - 1)2 = 64 (x + 3)2 + (y - 1)2 = 82 center: (-3, 1) radius = 8 (x + 3)2 + (y - 1)2 = 64.

  16. The general form of the equation of a circle is Ax2 + By2 + Dx + Ey + F = 0. Examples: General Form of the Equation of a Circle

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