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...

1 answer below »
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.
Y=-3x-6;(-1, 5) =
Write the equation of a line parallel to the given line but passing through the given point.



Y= -1/3X-4;(-6,-3) = Write the equation of a line perpendicular to the given line but passing through the given point.


This is an example of the format that is wanted


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 = -
?

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 -
?
.



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) this is the general form of the point-slope equation y – (-3) = -
?
[x



(-6)]



Substituting in my known slope and ordered pair y + 3 = -
?
x + (-
?
) 6



Simplifying double negatives and distributing the slope y = -
?
x



4



3



Because (-
?
)6 = -4 and 3 is subtracted from both sides


y = -
?
x



7 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) Substituting in the slope and ordered pair y – 5 = ¼ x The zero term disappears y = ¼ x + 5 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.]

Answered Same DayDec 29, 2021

Answer To: Given an equation of a line, find equations for lines parallel or perpendicular to it going through...

David answered on Dec 29 2021
112 Votes
Sol: (1)
The equation is given by 3 6y x   . The parallel line must pass through
point
 1,5 .
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 -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.
 1 1y y m x x   , this is the general form of the point-slope equation
 5 3 1y x      
Substituting in my known slope and ordered pair 5 3 3y x   
Simplifying...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here