Expert answer:Given an equation of a line, find equations for li

Answer & Explanation:PLEASE SEE ATTACHMENT FOR EXAMPLE ON HOW TO DO THIS DISCUSSIONGiven 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. y = -x + 3; (2, 2)
y = -x + 3; (2, 2)   ^Assigned problem to complete both steps.
Discuss the steps necessary to carry out each activity. Describe briefly
what each line looks like in relation to the original given line.
Answer these two questions briefly in your own words:

What does it mean for one line to be parallel to another?
What does it mean for one line to be perpendicular to another?

Incorporate the following five math vocabulary words into your discussion.
Use bold font to emphasize the words in your writing
(Do not write definitions for the words; use them appropriately in
sentences describing your math work.):

Origin
Ordered pair
X- or y-intercept
Slope
Reciprocal Attached is the example of this weeks discussion post, please make it like that but in your own words. mat221.w3.discussionexample.pdf
mat221.w3.discussionexample.pdf

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INSTRUCTOR GUIDANCE EXAMPLE: Week Three Discussion
Parallel and Perpendicular
For this week’s discussion I am going to find the equations of lines that are parallel or
perpendicular to the given lines and which are passing through the specified point. First I
will work on the equation for the parallel line.
The equation I am given is
The parallel line must pass through point
y = -⅔ x + 2
(-6, -3)
I have learned that a line parallel to another line has the same slope as the other line, so
now I know that the slope of my parallel line will be -⅔. Since I now have both the
slope and an ordered pair on the line, I am going to use the point-slope form of a linear
equation to write my new equation.
y – y1 = m(x – x1)
y – (-3) = -⅔[x – (-6)]
y + 3 = -⅔x + (-⅔)6
y = -⅔x – 4 – 3
y = -⅔x – 7
This is the general form of the point-slope equation
Substituting in my known slope and ordered pair
Simplifying double negatives and distributing the slope
Because (-⅔)6 = -4 and 3 is subtracted from both sides
The equation of my parallel line!
This line falls as you go from left to right across the graph of it, the y-intercept is 7 units
below the origin, and the x-intercept is 10.5 units to the left of the origin.
Now I am ready to write the equation of the perpendicular line.
The equation I am given is
y = -4x – 1
The perpendicular line must pass through point
(0, 5)
I have learned that a line perpendicular to another line has a slope which is the negative
reciprocal of the slope of the other line so the first thing I must do is find the negative
reciprocal of –4.
The reciprocal of -4 is -¼ , and the negative of that is –(-¼) = ¼. Now I know my slope
is ¼ and my given point is (0, 5). Again I will use the point-slope form of a linear
equation to write my new equation.
y – y1 = m(x – x1)
y – 5 = ¼ (x – 0)
y–5=¼x
y=¼x+5
Substituting in the slope and ordered pair
The zero term disappears
Adding 5 to both sides of the equation
The equation of my perpendicular line!
This line rises gently as you move from left to right across the graph. The y-intercept is
five units above the origin and the x-intercept is 20 units to the left of the origin.
[The answers to part d of the discussion will vary with students’ understanding.]
…
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