### Coordinate geometry

I will work problems. Hopefully that will give you some examples to follow.

Problem is to plot the following points (1, 0), (2, 3), (0, 4), (-2, 4), (-2, 0), (-2, -4), (0, -1) and (5, -3). The figure on the right shows those points plotted (roughly). Try to match the coordinate pairs with the x's on the figure. Remember all of the pairs are ordered pairs, meaning x followed by y. When you see points in (x,y ) it is almost always the x coordinate followed by the y coordinate.

Plot the points, find the distance between them and find the midpoint of the segment that joins them. The figure on the left gives you: a "rough" look at the points and the line that connects them. The calculations for determining the distance between the two points are also shown. The equation says that the straight line distance between any two points is the square root of the sum of the squared differences. This is Square Root of [(X2 - X1)2 + (Y2 - Y1)2.

To determine the midpoint of the line you will proceed as follows for the x coordinate:
• Xmid-pt = (X1 + X2)/2 = (2 + 4) /2 = 6/2 = 3.
• For the y coordinate the calculations are similar Ymid-pt = (Y1 + Y2)/2 = (-1 + 3)/2 = 2/2 = 1.
• So the coordinates of the mid-point are (3, 1).

Two questions. First sketch the region given by the set where y is greater than or equal to 0. This is all of the area above the x axis, including the x axis. Next sketch of the region where x is greater than 1 but less than 2 (not including 1 and 2). I have used dotted lines to indicate that the endpoints (1 and 2) are not included in the region.

Problem asks you to determine which point is closer to the point E (-2, 1). The two candidates are the points C (-6, 3) and D (3, 0). The technique we will use is to determine the distance from each point to E (-2, 1). The point with the smallest distance will be judged to be closer. To determine the distance we will use the distance formula discussed earlier.

Problem 40 - In this case you are given three points and asked to determine if they are on the graph of the equation. One way of determining this would be to plot the points and graph the equation and see if the points are on the graph. Another way would be to substitute the values of the point into the equation and determine if the result gives equality. This later approach is the the one we will use. In the work shown below each point is used in the equation. The first point, (1, 0) does not satisfy the equation so it would not be a point on its graph. The other two points do satisfy the equation so they would be points on the graph of the equation.

Problem is to: (1) make a table of values, (2) sketch the graph, (3) find the x and y intercepts and (4) test for symmetry. The equation given is y = 2x + 5. We will tackle them in order.

The table of values

X ValueY Value
-3-1
-21
-13
-05
17
29
311

The figure

Intercepts and symmetry

The problem asks us to show that the given equation represents a circle.

x2 + y2 - 4x + 10y + 13 = 0

• First we will group terms by variable
• x2 - 4x + y2 + 10y + 13 = 0
• Now we will complete the square for each variable
• For x 4/2 = 2 and 2 squared is 4, for y 10/2 = 5 and 5 squared is 25
• x2 - 4x + 4 -4 + y2 + 10y + 25 - 25 + 13 = 0
• (x - 2)2 - 4 + (y + 5)2 - 25 + 13 = 0
• Group the constant terms and moving the net value to the right side of the equation
• (x - 2)2 + (y + 5)2 = 16 (or 42)
• This is a circle with center at (2, -5) and radius equal to 4