Topic 3 - (3.1)  -  Introduction to Graphing

Plot the following points and tell what quadrant they are in.

  (-2, 1)   Start at the origin, move two units to the left and from that point 1 unit up.

              The point is in quadrant II.

  (-3, -5)  Start at the origin, move three units to the left and 5 units down.

               The point is in quadrant III.

  (4, 0)     Start at the origin, move 4 units to the right and stop.

               This point isn't in any quadrant, it is an x-intercept.

  (0, -2)    Start at the origin, move down 2 units.

               This point isn't in any quadrant, it is a y-intercept.

  (4, -1)    Start at the origin, move 4 units to the right and down 1.

               This point is in quadrant IV.

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Find the rise and run in moving from the point (2, 5) to (11, 9).

Rise:  | y₂  -  y₁|  = | 9  -  5 |  =  | 4 |  =  4   The rise is the absolute value of the

                                                               difference between the y value of the

                                                               second point and the y value of the

                                                               first point.  (distance always positive)

Run:  | x₂  -   x₁ |  =  | 11  -  2 |  =  | 9 |  =  9  The run is the absolute value of the

                                                                   difference between the x value of the

                                                                   second point and the x value of the

                                                                   first point.

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Given P₁(8, 9), find the coordinates of P₂ if the rise from P₁ to P₂ is 4 and the run is 7.

P₁ (8, 9)       P₂(x₂, y₂)                 Use the rise and run ideas from the last problem.

rise  =  | y₂ - y₁ |                   run  =  | x₂ - x₁ |

  4   =  | y₂ - 9 |                       7   =  | x₂ - 8 |

 4  =  y₂ - 9  or  -4  =  y₂ - 9     7  =  x₂ - 8  or  -7  =  x₂ - 8

 4 + 9  = y₂    or  -4 + 9 = y₂     7 + 8 = x₂  or  -7 + 8 = x₂

   13  =  y₂                5  =  y₂          15  =  x₂            1  =  x₂  

There are several possible points, but when you start at (8, 9) and the run is 7 (to the right) and the rise is 4 (up).  The answer has to be     (15. 13)

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If a  =  15 and b  =  36, use the Pythagorean Theorem to find c.

c is always the hypotenuse of the right triangle in these problems.  a & b are legs.

The Pythagorean Theorem:     a²  +  b²  =  c²     a  =  15    b  =  36

        a²     +     b²     =     c²             Substitute values for a & b.

      (15)²   +    (36)²  =     c²             Perform the operations and solve for c.

     225      +    1296  =     c²

              1521           =     c²            Take the square root of both sides and only

                 39           +      c               use the positive. (sides of triangles are always

                                                       positive.)

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What is the equation of the circle whose center is at  (-3, 1) and has radius 57.

         (x  +  3)²  +  (y  -  1)² = (57)²