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## MATH121, 3.4 Notes

by: Mallory McClurg

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# MATH121, 3.4 Notes Math 121

Mallory McClurg
OleMiss
GPA 3.37

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These notes cover all of Chapter 3.4 - Parallel and Perpendicular Lines. Step-by-step explanations of every type of problem we'll be tested on. Taken right from my "certify" homework!
COURSE
College Algebra
PROF.
Dirle
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
Math121, parallel, perpendicular
KARMA
25 ?

## Popular in Math

This 5 page Class Notes was uploaded by Mallory McClurg on Tuesday September 20, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 4 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/20/16
Math121 Chapter 3 Lesson 3.4 – Parallel and Perpendicular Lines EXAMPLE 1. 4x + 2y = 10 (First, we’re asked to put this in y = mx + b form.) 2y = -4x + 10 y = -2x + 5 (Now, we’re asked to find the equation for a line that is parallel to the line made by the linear equation we just got… and it runs through the point (8, 2). So, take the x and y from the point we were given, and…substitute it in to the linear equation we wrote. Remember, a parallel line means that it will have the exact same slope as the original line.) 2 = -2 (8) + b (Now, solve for b.) b = 18 (Now that we have b, we can substitute this into a new equation for the line parallel to the one in the original equation, using the same slope.) y = -2x + 18 (This is the linear equation to the line parallel to the original!) EXAMPLE 2. 2(y – 1) + ((4x + 5)/3) = -4 (First, we’re asked to put this in slope- intercept form, which is y = mx + b form. But we need to get y on a side by itself first, so…to get rid of the fraction, multiply everything by 3.) 6(y – 1) + (4x + 5) = -12 (Distribute the 6 to the y and -1.) 6y – 6 + 4x + 5 = -12 (Get the y on a side by itself.) 6y = -4x – 11 y = (-2/3)x – (11/6) (This is in proper form. Now, the question asks us to find the equation of a line parallel to this one, that runs through the point (11, 12), so substitute the x and y into the linear equation we found, and use the same slope, since it is parallel.) 12 = (-2/3)(11) + b (Now, solve for b.) b = (58/3) (Now, write out a new linear equation in y = mx + b form and substitute in the new value for b.) y = (-2/3)x + (58/3) (This is the linear equation for the parallel line.) Math121 Chapter 3 EXAMPLE 3. 3 – ((6y + 5x)/3) = 1 (To get this in y = mx + b form subtract 3 from both sides of the equation.) ((6y + 5x)/-3) = -2 (Now, multiply both sides of the equation by -3 to get rid of the fraction.) 6y + 5x = 6 (Get y on a side by itself, so this is in proper form.) y = (-5/6)x + 1 (We’re asked to find the equation for a line parallel to this line that runs through the point (13, -10), so substitute 13 for x and -10 for y in the equation we just found.) -10 = (-5/6)(13) + b b = (5/6) (Now, use this new value for b, and use the same slope as the original, because it’s parallel.) y = (-5/6)x + (5/6) (This is our parallel line!) EXAMPLE 4. 3x + 4y = 3y – 2 (First, get y on a side by itself.) y = -3x – 2 (Now, we’re asked to find the equation for a line that is perpendicular to the original and runs through the point (13, -10). With parallel lines, we use the same slope, but with perpendicular lines, all we have to do is get the negative inverse of the slope, so the slope of the perpendicular line is (1/3). Substitute the x and y values from the point, into the problem, and use the perpendicular slope.) -10 = (1/3)(13) + b (Solve for b.) b = (-43/3) (Now, using the perpendicular slope, substitute this b into the y = mx + b equation.) y = (1/3)x – (43/3) (This is the equation for the perpendicular line!) Math121 Chapter 3 EXAMPLE 5. ((3x + 2)/3) – 2y = 1 – y (First, we need to get this in proper form. First, multiply everything by 3 to get rid of the fraction.) 3x + 2 – 6y = 3 – 3y (Now, get y on a side by itself.) -3y = -3x + 1 y = x – (1/3) (This is the linear form of the equation! Now, the question asks us to find the line that passes through the point (14, 5) and is perpendicular to our original line, so our new slope is -1 since that is the negative inverse of 1 (the coefficient of x). We should substitute the new slope, and the x and y values in the given point into the y = mx + b equation and then solve for b.) 5 = -1(14) + b 19 = b (Now, with our new slope, substitute this value for b into the y = mx + b equation.) y = -x + 19 (This is the perpendicular line!) EXAMPLE 6. ((5x – 2y)/2) = ((x – 4)/4) (To get this into proper form, we need to multiply both fractions by 4 to get rid of the fractions. Then, simplify the fractions.) 2(5x – 2y) = x – 4 (Distribute the 2 to the 5x and 2y.) 10x – 4y = x – 4 (Get y on a side by itself.) -4y = -9x – 4 y = (9/4)x + 1 (Now, the question wants us to find the equation to a line that passes through (-12, -8) and is perpendicular to this line we just found, so substitute the x and y values for the given point with the negative inverse of the slope in the original problem.) -8 = (-4/9)(-12) + b (Solve for b.) Math121 Chapter 3 b = (-40/3) (With the new slope, substitute the new value for b into the y = mx + b equation.) y = (-4/9)x – (40/3) (This is the equation for the line perpendicular to the original!) EXAMPLE 7. 2 – ((3y + 2x)/3) = 3 (First, we need to get this in proper form. Subtract 2 from both sides. If you stick the minus sign with the 3 in the denominator of the fraction, this problem will be a lot easier!) ((3y + 2x)/-3) = 1 (Multiply both sides by -3.) 3y + 2x = -3 (Get y on a side by itself.) y = (-2/3)x – 1 (The question now asks us to find the parallel line that passes through the point (14, 5), so substitute the point values for x and y, and use the same slope because the line needs to be parallel!) 5 = (-2/3)(14) + b (Solve for b.) b = (43/3) (Substitute this new value for b into the equation y = mx + b with the same slope as the original problem.) y = (-2/3)x + (43/3) (This is the equation for the perpendicular line!) EXAMPLE 8. ((3x + 5)/3) – 5y = 2 – 3y (We need to get this in proper form, so we should first multiply everything in the equation by 3 to get rid of the fraction.) 3x + 5 – 15y = 6 – 9y (Get y on a side by itself.) -6y = -3x + 1 y = (1/2)x – (1/6) (This is the simplified linear equation. Now, we’re asked to find the equation of a line perpendicular to this, that passes through the point (-12, -2), so substitute everything given into the y = mx + b equation. The perpendicular slope, which is the negative inverse of the original slope, is -2.) Math121 Chapter 3 -2 = -2(-12) + b b = -26 (Substitute this new value for b into y = mx + b with the perpendicular slope.) y = -2x – 26 (This is the linear equation for the line perpendicular to the original.)

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