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Question 1186326: Find the equation of the line parallel to 3x + 4y = 20 and at a distance of 5 units from this
line.
Found 3 solutions by MathLover1, Boreal, greenestamps:
Answer by MathLover1(20849)   (Show Source):
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! 3x+4y=20
4y=-3x+20
y=-(3/4)x+5
graph this
Find the line perpendicular to this line. It will have a slope of (4/3), the negative reciprocal.
The simplest such line is y=(4/3)x, that goes through the origin.
The two lines intersect where -(3/4)x+5=(4/3)x
(25/12)x=5
25x=60
x=2.4
(4/3)x=3.2, so the point is (2.4, 3.2), which is a solution to the given line as well
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5 units away along the perpendicular line will change the x (4/3) as much as the y. Think of that as the hypotenuse of a triangle with legs 3 and 4. The hypotenuse will be length 5, which is what is wanted. The x goes down or up 3 from any point on the line, and the y goes down or up (same diirection as x since slope is positive) 4 units.
Since (0, 5) is a point on the original line, a point (3, 9) would be a point on the line desired. Because the line desired is parallel to the original line, it has slope -3/4.
Point slope formula y-y1 = m(x-x1), m slope, (x1, y1) point
y-9=(-3/4)(x-3)
y=(-3/4)x+11.15
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Go the other way and (-3, 1) is a point
y-1=(-3/4)(x+3)
y=(-3/4)x-1.25
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
The solution from tutor @MathLover 1 is simply wrong. Lines parallel to the given line with y-intercepts that are 5 more or less than the y-intercept of the given line are NOT 5 units from the given line.
The solution from tutor @Boreal is fine, except for a typo on one of the equation he shows as the answers.
Here is what I think is an easier way to get to the answers, using the basic idea of @Boreal's solution but in a different way.
The y-intercept of the given line is (0,5); its slope is -3/4.
To find the equations of the two lines parallel to the given line and a distance of 5 from the given line, do the following:
(1) Pick a convenient point A on the given line; the obvious convenient point is the y-intercept of the given line, (0,5)
(2) On the line perpendicular to the given line at (0,5), find points B and C that are a distance of 5 from (0,5).
The slope of the given line is -3/4; the slope of the line perpendicular to the given line is 4/3. Knowing that 3-4-5 is a Pythagorean Triple, to find a point 5 units from (0,5) on a line with slope 4/3, we can either (a) move 3 units right and 4 units up to reach B(3,9) or (b) move 3 units left and 4 units down to reach C(-3,1).
(3) Find the equations of the two lines parallel to the given line and 5 units from the given line, knowing that any line parallel to the given line will have an equation of the form 3x+4y=C. Plug in the values of the points B and C to find the equations we are looking for.
through B(3,9):  --> 
through C(-3,1):  --> 
ANSWERS:
(1)  , or  , or 
(2)  , or  , or 
And here is an easier way to find the answers....
Given a point (m,n) and a line with equation Ax+By+C=0, the distance from the point to the line is
In this problem, the equation of the given line -- in the form required for using the formula -- is 3x+4y-20=0; a convenient point on the given line is (0,5), and the desired distance is 5:
 or 
 or 
That gives us the same two equations we found earlier:
 and 
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