•Read the following instructions in order to complete this discussion, and review the example of how to complete the math required for this assignment:

◦Given an equation of a line, find equations for lines parallel or perpendicular to it going through specified points. Find the appropriate equations and points from the table below. Simplify your equations into slope-intercept form.

◦Use your assigned number to complete.

8                                                                 y = 3x + 3; (1, 1)                                                      y = 3x + 3; (1, 1)

Your initial post should be 150-250 words in length. Respond to at least two of your classmates’ posts by Day 7 in at least a paragraph. Make sure you choose people who don’t have the same equations as you worked. Do you agree with how they used the vocabulary? Do their equations seem reasonable given what they started with?

Carefully review the Grading Rubricfor the criteria that will be used to evaluate your discussion.

Response 1= Jason Sessions

Hello Class,

                Today, we are going to find a parallel and perpendicular lines. First, we must understand the difference between these two lines. A parallel line runs next to another, and never crosses each other. Whereas, a perpendicular will have a path which runs in the opposite direction and crosses the other line. Example of a parallel would be the equals sign “=” and a perpendicular would be the plus sign “+”.

                Now we understand the difference, let us look at the problem we are to solve today.

y=2x with a pass-through point being (-3, -3)

Our pass-through point in one of the points where the x and y intercept meet on the line. In this case, if we were to graph these point, they would meet to the left on the x intercept three points from the origin point of (0,0), and down three points of the y intercept from the origin point. The ordered pair for the x and y intercept is written like (x, y).

                First, we will find a line which runs parallel to y=2x and runs through the grid point of (-3, -3).

y=2x                            To find the parallel, we need to solve the equation y=mx+b. y=-3, x=-3, and m=slope=2 in this case.

-3=2(-3)+b                  We fill in the known variables. Y is -3 and our slope is giving in the original problem as 2, and x is -3 as well.

-3=-6+b                       We solve for 2(-3), which is 2*-3=-6

+6-3=-6+6+b              To get “b” by itself, we need to add “6” to both sides.

3=b                              “b” is equal to 3.

{y=2x+3}                    This is our equation for the parallel line of y=2x, and our slope is 2.

                Next, we will find the perpendicular line to y=2x with the pass-through points remaining (-3, -3).

y=2x                            We are going to use the same equation to find the perpendicular line. However, you will notice an added step.       

-3=2(-3)+b                  We filled in the variables.

-3=-1/2(-3)+b             The perpendicular slope is the reciprocal of the original slope. In this case, -1/2 is the reciprocal of 2.

-3=3/2+b

-3/2-3=3/2-3/2+b

-3/2-6/2=b                   We need to have both denominators the same. So we -3*2 and -1*2 which gives us -6/2.

-9/2=b

{y=-1/2x-9/2} Jason

Response 2= Ashley Jackson

Original equations:

Parallel = y= – 2/3 x + 2 (6,3) to find the parallel equation, first you must find the slope which the original rule for slope is y = mx + b. The slope is the m in the equation. So the slope in this equation is -2/3. Then plug in 6 for y and 3 for x to solve and get the new equation for the parallel line.

Perpendicular — y= – 2/3 x + 2 (6,3) to find the perpendicular equation, find the slope. The slope for this is a little different because the perpendicular line makes a right angle with the original line. To find the slope of the new line you must make the two slopes when multiplied equal -1. -2/3 multiplied by 3/2 equals -1. So the slope for this is 3/2. Then plug 6 for x and 3 for y to solve and get the full new equation for the perpendicular line.

New Solved equations:

Parallel — y= -2/3 x + 7 [line crosses through positive y axis at (0,7) then continues down and to the right then passes through the positive x axis at approximately (10,0)]

Perpendicular — y= 3/2 x – 6 [line crosses the negative x axis at (-4,0) then continues down and to the right then crosses the negative side of the y axis at (0,-6)]

The ordered pairs I gave, in the explanation of what the new lines look like, are the x-intercept and y-intercept of the new lines. The x-intercept is where the line crosses the x axis, and the y-intercept is where the line crosses the y axis in relation to the origin of the graph

Parallel line – two lines that never meet and are always the same distance apart, like a train track. They never meet and are always the same distance apart.

Perpendicular line – two lines that are right angles (90 degrees) to each other, creating a perfect X or cross out of the two lines.

If the original equations had whole numbers rather than fractions then to find the perpendicular line’s equation, you would find the reciprocal to create the “rise over run” for the slope.

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