# Thread:

1. ## Finding The Distance Between Parallel Lines...?

How do you find the distance between the two parallel lines:

y = 4x+6
y = 4x-9

A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...

Is anyone smart enough to do this question?

2. Originally Posted by AlphaRock
How do you find the distance between the two parallel lines:

y = 4x+6
y = 4x-9

A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...

Is anyone smart enough to do this question?
$0=4x+6\Rightarrow{x=\frac{-3}{2}}$
$0=4x-9\Rightarrow{x=\frac{9}{2}}$
$\frac{3}{2}+\frac{9}{2}=3.75$

3. Originally Posted by Mathstud28
$0=4x+6\Rightarrow{x=\frac{-3}{2}}$
$0=4x-9\Rightarrow{x=\frac{9}{2}}$
$\frac{3}{2}+\frac{9}{2}=3.75$
I also thought it'd be 3.75, but suprisingly, it's 3.64.

My friend explained this to me hour ago, but I accidentally exited the conversation.

4. Originally Posted by AlphaRock
I also thought it'd be 3.75, but suprisingly, it's 3.64.

My friend explained this to me hour ago, but I accidentally exited the conversation.

It is 3.75...I think your book may be wrong

5. Originally Posted by Mathstud28
It is 3.75...I think your book may be wrong
Probably, but if it helps... it included tangent that opp is 4 times bigger than four or something. And that the tangent = slope.

and x^2 + #x^2 = 225
17x^2 = 225
4.12x = 15
x = 3.64

These are all that I can remember. I can't remember the steps or understanding in between. Perhaps you can.

6. I REALLY don't know where the X^2 or 17 came from. If somebody could explain everything clearly... I would appreciate it.

7. What Mathstud found was the horizontal distance between two points of the two lines which isn't necessarily the shortest distance.

Recall the distance formula from a point to the line (presumably the shortest):
$d = \frac{|Ax_{0} + By_{0} + C|}{\sqrt{A^{2} + B^{2}}}$

where (x_{0},y_{0}) is your point and Ax + By + C = 0 is your line.

Now, let's pick a point from y = 4x + 6 ... let's say (0,6). This will be our point.

From our other line y = 4x - 9, we must but it in standard form (Ax + By + C = 0):
$y = 4x - 9 \: \Rightarrow \: 0 = 4x - y - 9$

Now apply the formula:
$d = \frac{|4(0) + (-1)(6) + (-9)|}{\sqrt{(4)^{2} + (-1)^{2}}} \approx 3.638$

8. Also, note that the shortest distance between two lines is a perpendicular line that intersects both.

Start with y = 4x + 6. Let's use the point (0,6). We want to find the equation of a line perpendicular to y = 4x + 6 that goes through (0,6). Perpendicular lines have negative reciprocal slopes, so our line will have a slope of -1/4 - note that it has the same y-intercept.

So our equation is y = (-1/4)x + 6. To see where that line intersects our OTHER line (in other words, to find the point on the other line closest to (0,6), set the two equal to one another.

(-1/4)x + 6 = 4x - 9. Solve for x. It isn't pretty, but you get x = 60/17. Plug that in to either our perpendicular line's equation or the second equation to get y, which is 87/17. So now we have these two points, (0,6) and (60/17, 87/17), and we know, since they're on a perpendicular line that intersects our two original lines, that they're as close as possible. But how close?

The distance formula:

${\sqrt{\frac{60}{17}^2 + \frac{15}{17}^2}\approx 3.638}$

Admittedly, this is a longer method than the one o_O proposed, but I never remember the equation for the distance between a point and a line - this particular method relies on a little more "basic" geometrical concepts and formulas. And, having multiple methods to solve a problem is never a bad thing.

9. The triangles in my diagram are similar.

Using this fact you can label one triangle with sides 15, 4x and x and then use Pythagoras' theorem.

$x^2+(4x)^2=15^2$

$17x^2=15^2$

$x=\frac{15}{\sqrt{17}}$

$x=3.638..$

10. Does this apply to any dimension.if we have a1*x1+a2*x2+...an*xn=b1 and a1*x1+a2*x2+...an*xn=b2.

11. ## Yes

Yes, the formula works for higher dimensions as well

If we have a hyperplane wx=b, then the distance from point x1 (which is a vector) to the line is given by:

|wx-b|/||w||^2

Remember w is also a vector of weights so ||w||= w1^2+w2^2+.....
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