How to

MAT101, SLP1

Revised March 2013 (Dr. Rensvold)

INSTRUCTIONS: Read the references given under Background Information. Also view the instructional videos under the following TD2 topics: "Using the Equation Editor" and "Plotting Equations." Work the following problems, taking full advantage of the hints, and showing all your work.

1. If the point (2,5) is shifted 3 units to the right and 2 units down in the X-Y plane, what are its new coordinates? (Hint: Movement right/left = +/- X; Movement up/down = +/- Y.)

2. If the point (-1,6) is shifted 2 units to the left and 4 units up, what are its new coordinates? ANS: distance^2 = (x1 - x2)^2 + (y1 - y2)^2

hence with this formula, we can find y.

13^2 = [3 - (-2)]^2 + [-(1) - y)]^2

169 = 25 + 1 + 2y + y^2

y^2 + 2y - 143 = 0

(y-11)(y+13) = 0

y = 11, y = -13

3. Find all the points having an x-coordinate of 3 whose distance from the point (-2, -1) is 13. (Hint: The unknown points [there are two of them] have coordinates (3,y), where y is to be determined. Use the coordinates of the given point, the distance, and the distance formula to solve for the required values of y.) ANS” For two points, (x1, y1) and (x2, y2), the distance equation is :

d = √[x1 - x2)^2 + (y1 - y2)^2]

Let (x1, y1) be (3, y) and let (x2, y2) be (-2, -1) and let d = 13.

Then, on substitution,

13 = √ { [3 - (-2)]^2 + [y - (-1)]^2 }

Simplifying :

13 = √ [25 + (y + 1)^2]

Square both sides :

169 = 25 + (y + 1)^2

Rearrange :

(y + 1)^2 = 169 - 25 = 144

Take square root of both sides :

y + 1 = ± 12

Therefore, y = 12 - 1 = 11,

or, y = -12 - 1 = -13.

Thus, there are the 2 points (3, 11) and (3, -13).

4. Find all points on the y-axis that are 6 units from the point (4, -3).

ANS: Let (0, k) be the point on the y-axis.

Using the distance formula:

= 6

= 6, square both sides

= 36

= 20, get the square root of both sides

k + 3 = ±

k = -3 ±

The points are (0, -3 - ) and (0, -3 + ).

5. Graph the line...