algebra parallel and perpendicular

algebra parallel and perpendicular

Slope

Reciprocal
Your initial post should be 150-200 words
Example how to complete assignment by instructor:
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 y = -2/3 x + 2 The parallel line must pass through point (-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 -2/3. 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) = -2/3[x – (-6)] y + 3 = -2/3x + (-2/3)6
y = -2/3x – 4 – 3
y = -2/3x – 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 (-2/3)6 = -4 and 3 is subtracted from both sides The equation of my parallel line!
left to right across the graph of it, the y-intercept is 7 units
This line falls as you go from
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 -1/4 , and the negative of that is –(-1/4) = 1/4. Now I know my slope is 1/4 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 = 1/4 (x – 0) y – 5 = 1/4 x
y = 1/4 x + 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.]
ORDER THIS ESSAY HERE NOW AND GET A DISCOUNT !!!

You can place an order similar to this with us. You are assured of an authentic custom paper delivered within the given deadline besides our 24/7 customer support all through.