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Bunuel
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Difference between the y intercepts is not equal to the distance between 2 lines except when both lines are parallel to x-axis.

Suppose angle ABC = x

Difference between y-intercepts= AB

Distance between the 2 lines= AC

AB*sin(x)= AC


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Bunuel
The distance between two parallel lines given by equation \(x + 2y - 3 = 0\) and \(2x + 4y - 17 = 0\) is approximately:

A. 0.85
B. 1.25
C. 1.95
D. 2.5
E. 3.25


Are You Up For the Challenge: 700 Level Questions

transform the equations into slope intercept form
1)x + 2y - 3 =0 -> y= -(x/2) + 3/2
2)2x + 4y - 17 = 0 -> y = -(x/2) + 17/4

Since the lines are parallel, you can simply take the difference between the y intercepts to get the distance between the lines

Distance = 17/4 - 3/2 = 11/4 ≈ 2.5.

Also, one could intuitively understand that 11/4 must be greater than 2 and lesser than 3, answer choice (D) -> 2.5

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nick1816
Difference between the y intercepts is not equal to the distance between 2 lines except when both lines are parallel to x-axis.

Suppose angle ABC = x

Difference between y-intercepts= AB

Distance between the 2 lines= AC

AB*sin(x)= AC


ShreyasJavahar
Bunuel
The distance between two parallel lines given by equation \(x + 2y - 3 = 0\) and \(2x + 4y - 17 = 0\) is approximately:

A. 0.85
B. 1.25
C. 1.95
D. 2.5
E. 3.25


Are You Up For the Challenge: 700 Level Questions

transform the equations into slope intercept form
1)x + 2y - 3 =0 -> y= -(x/2) + 3/2
2)2x + 4y - 17 = 0 -> y = -(x/2) + 17/4

Since the lines are parallel, you can simply take the difference between the y intercepts to get the distance between the lines

Distance = 17/4 - 3/2 = 11/4 ≈ 2.5.

Also, one could intuitively understand that 11/4 must be greater than 2 and lesser than 3, answer choice (D) -> 2.5

Ahh, you're right, my bad. Thanks for pointing it out.
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The equations given are
1)x + 2y - 3 =0
2)2x + 4y - 17 = 0

The same equations in y=mx + c form gives
1) y= -(x/2) + 3/2
2) y = -(x/2) + 17/4

Now both lines being parallel have same slopes.
Also for distance between two lines, the distance between y intercepts is what we need to calculate.
Hence y intercept putting x=0 gives 3/2 and 17/4.
Hence Distance = 17/4 - 3/2 = 11/4 or 2.5

Answer choice is (D)
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Bunuel
The distance between two parallel lines given by equation \(x + 2y - 3 = 0\) and \(2x + 4y - 17 = 0\) is approximately:

A. 0.85
B. 1.25
C. 1.95
D. 2.5
E. 3.25

Distance Parallel Lines: \(\frac{|c_2-c_1|}{\sqrt{Aˆ2+Bˆ2}}\)

Line form: \(Ax+By+C\) where A,B_1 and A,B_2 must be equal

\(x + 2y - 3 = 0…2x+4y-6\) and \(2x + 4y - 17 = 0\)

A=2, B=4, c_1=-6, c_2=-17

\(\frac{|c_2-c_1|}{\sqrt{Aˆ2+Bˆ2}}=\frac{|-17-(-6)|}{\sqrt{2ˆ2+4ˆ2}}=\frac{11}{\sqrt{20}}=aprox2.5\)

Ans (D)
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Bunuel
The distance between two parallel lines given by equation \(x + 2y - 3 = 0\) and \(2x + 4y - 17 = 0\) is approximately:

A. 0.85
B. 1.25
C. 1.95
D. 2.5
E. 3.25


Are You Up For the Challenge: 700 Level Questions

I do not like to use formulas when you can use other method, so basically what I did is draw both lines, find a perpendicular line to both lines, and with equations, points and slope I was able to find a new point at the intersection and then apply the formula for distante of 2 points. See the attachment with the solution.
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