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 2 answersLast reply by: Jing ChenTue Aug 29, 2017 2:02 PMPost by Sai Nettyam on October 11, 2011Ms. Pyo,How do you find the distance between equations? For example... Find the distance between the lines with the equations y= 2/7x + 4 and y= 2/7x -2. Can you please explain on how to solve problems like this. Thank you.

### Parallels and Distance

• The distance from a line to a point not on the line is the length of the segment perpendicular to the line from the point
• the distance between two parallel lines is the distance between one of the lines and any point on the other line

### Parallels and Distance

Draw the segment that represents the distance from point A to MN .

AB represents the distance.
Draw the segment that represents the distance from point A to MN .

AB represents the distance.
Determin whether the following statement is true or false.
Line m is parallel to line n, point A is on line m, the distance from point A to line n is d, then the ditance between lines m and n is d.
True.
Draw the segment that represents the distance from point A to MN .

AB represents the distance.
Line p is parallel to line q. Draw a segment represents the distance between line p and line q.
• Find point A on line p
• Draw AB ⊥ line q.

AB represents the distance.
Line m intersects line n at point A(6, 9), line m^line n, point B(3, 7) is on line m, find the distance from point B to line n.
• AB represents the distance between line m and line n.
d = √{(6 − 3)2 + (9 − 7)2} = √{9 + 4} = √{13} .
Line p is parallel to line q. Draw the segment that represents the distance from point A to line q.

AB represents the distance.
Line p is parallel to line q, the distance between the two lines is 2x + 4, point A is between two lines, the distance from point A to line p is x + 4, the distance from point A to line q is 3, find x.
• 2x + 4 = (x + 4) + 3
x = 3
The given equation represents line m and point A(3, 4). Construct the perpendicular segment and find the distance from the point to the line.
y = − 3x + 2
• Assume line n passes through point A and is perpendicular to line m
• the equation for a line m is y = kx + b
• line m ^line n, so k = [1/3]
• line n passes through point A
• 4 = [1/3]*3 + b
• b = 3
• so line n is y = [1/3]x + 3
• Assume that line m and line n intersect at point B, − 3x + 2 = [1/3]x + 3
• x = − [3/10]
• y = − 3*( − [3/10]) + 2 = [29/10]
• B( − [3/10], [29/10])
• AB represents the distance between point A and line m
• d = √{(3 − ( − [3/10]))2 + (4 − [29/10])2}
d = [(√{1110} )/10].
Draw the segment that represents the distance from point A to MN

AM represents the distance

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Parallels and Distance

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Distance Between a Points and Line 0:07
• Definition and Example
• Distance Between Parallel Lines 1:51
• Definition and Example
• Extra Example 1: Drawing a Segment to Represent Distance 3:02
• Extra Example 2: Drawing a Segment to Represent Distance 4:27
• Extra Example 3: Graph, Plot, and Construct a Perpendicular Segment 5:13
• Extra Example 4: Distance Between Two Parallel Lines 15:37

### Transcription: Parallels and Distance

Welcome back to Educator.com.0000

This next lesson is on parallel lines and distance.0002

The first one: the distance between a point and a line: we have a point here, and we have a line.0010

And to find the distance between them, you have to find the length of the segment that is perpendicular.0017

If you were to get a ruler, you need to find the distance--how far away this point is from this line.0026

You are going to take your ruler, and you are going to make it so that the ruler is going to be perpendicular to the line.0035

So, imagine if that point is you; you are that point; you are standing in the room, and this line is the wall in front of you.0043

So, if you have to find the distance between you (this point right here) and the wall (the line), then you don't find the distance this way.0056

You don't go diagonally to the wall; if you are facing directly on the wall, you have to measure the distance0067

so that your pathway, that length of that segment, is going to be perpendicular to that wall.0074

If you are going to find the distance, you are not going to do this; that is not the distance.0090

It has to be from the point to the line so that it is perpendicular, so this right here will be the distance.0098

Now, to find the distance between two parallel lines: we know that parallel lines never intersect.0112

They are always going to have the same distance between them, no matter where you measure it from.0119

Now, that word right there is called "equidistant"; that means that, no matter where you measure the distance, it is always going to be the same.0126

And that is equidistant--an "equal distance," so "equidistant."0138

And so then, that means that, if you want to find the distance between them, you can just find it for any point on the line,0145

and make sure, just like we just went over, that the point and the line are perpendicular.0155

This segment right here: to find the distance, they have to be perpendicular to each other.0163

This would be the distance; if you look for it right here, as long as it is perpendicular, that will also be the distance.0168

That is the distance between two parallel lines.0179

Let's go over a few examples: Draw the segment that represents the distance from point A to EF.0187

And this is segment EF; so here is point A; here is segment EF.0195

I want to draw the segment that represents the distance.0205

If I go like this, the distance this way to the line, that is going to be incorrect, because that is not perpendicular.0212

This right here, what I drew (it is supposed to be a segment)--that is not perpendicular.0225

What you have to do: you have to extend this out; make sure that it is lined up, extend it out...and then you would draw that.0231

This is your distance, and the same thing here: we are going to extend this out, and then draw the segment that is perpendicular to it.0247

And you measure it, and that is the distance.0262

Let's do a couple more: here is point A; here are two parallel lines and point A; you want to find the distance from this point to this line.0267

So, now, if I go like this, is that the distance? No, this is not the distance.0277

You have to make sure to draw it like this, so that it is perpendicular; and that would be the distance.0288

Here is point A; here is segment EF; in this case, you would draw it; that is perpendicular.0297

Graph the given equation (there is your equation) and plot the point; construct a perpendicular segment, and find the distance from the point to the line.0315

So, first, let's draw the line; y = 2x - 1: this is y = mx + b; b is your y-intercept; that means I have to go to -1 and plot that as my y-intercept.0324

Here is my x; here is my y; and then, my slope is 2/1--remember, this is rise over run.0344

If my rise is 2, and it is positive 2, that means that I am going to go up 2, and then I am going to move to the right 1, because that is a positive.0356

For positive, you go right; for negative, you go left.0367

And then, you are going to go again: 1, 2, and you can just keep doing that.0370

You could go from this point--go to -2 and -1, and that is still going to give you a positive 2.0373

There is my line, right there; and then, the point is 0, 1, 2, 3.0381

First, we have graphed the equation; we have plotted the point; construct a perpendicular segment, and find the distance.0395

If we know that this right here, the slope of this line, was 2/1, that means that0407

if I draw a perpendicular segment from this point, the slope is going to be the negative reciprocal.0419

Remember: if they are perpendicular lines, then the slope is going to be the negative reciprocal.0429

For this other line, the perpendicular line, my slope will be -1/2.0438

Now, this can either mean -1 over 2, or 1 over -2; it is the same thing.0447

So, if I make it -1 over 2, from here, -1 is my rise, over run; so for the rise, since it is -1, I am going to go down 1, and 1, 2--over 2.0454

Down 1, and over 2, down 1 and over 2...and then, it is going to look something like this.0471

Now, to find the distance from that point to this point on the line, first of all, this is an intersection point between these two lines;0493

so, how can I find the intersection point between these two lines?0503

Well, I have my equation here; this is my first equation; my second equation, right here, is just going to be y =...0509

my slope was -1/2x; my y-intercept was 3; so all I did was to plug in my slope and my y-intercept.0525

The slope is -1/2; the y-intercept is 3; and then, you can just solve those two out, so y = 2x - 1.0536

I am going to use the substitution method...plus 3...equals 2x - 1.0547

And then, if I subtract the 2x over to this side, this is going to be -5/2x is equal to...0555

I am going to subtract the 3 over there, so I am going to get a 4.0566

Now, all I am doing is solving this out; remember, this is systems.0569

And then, to solve for x, I need to multiply this whole thing by the reciprocal, so -2/5.0575

That way, this is going to all cancel and give me 1; multiply that side by it; so x is going to equal 8/5--there is my x, and then my y.0584

I just need to plug it back into one of these; it doesn't matter which one; let's plug it into this one.0606

2 times x is 16/5 - 1; so this will be y = 2(8/5) - 1; here is 16/5 - 1, or I can make this 5/5, and that is going to be 11/5.0610

So, my point is this point right here; it is 8/5 and 11/5.0633

Now, the whole point of doing that (let me just explain), of finding this point, is so that you can find the distance between the two points.0650

If you have this point, and you have this point, you can use the distance formula to find how long the segment is.0662

All of this work was just to find this point right here, because that is the point where they intersected.0671

So then, I am going to use those two points; I am trying to find the distance between (0,3) and (8/5,11/5).0679

The distance formula is the square root of (x2 - x1)2 + (y2 - y1)2.0700

The distance is going to be (0 - 8/5)2 + (3 - 11/5)2.0712

This is -8/5; that is 64/25, plus...3 - 11/5; if you want to subtract these, you need a common denominator.0733

My common denominator is going to be 5; that is 15/5 - 11/5.0747

So, this is going to be 4/5, and that is squared; so it is going to be 16/25.0755

And then, the square root of...I have a common denominator, so this is going to be...80/25 (let me make some room over here):0775

that is the same thing as the square root of 80 over the square root of 25.0799

The square root of 80: this is 8 times 10; 8 is 4 times 2...let me just show you a quick way;0803

I know this is a little off of this lesson, but to simplify square roots, if you have √80, and you want to simplify it,0820

then you can just do the factor tree: this is 8 and 10; this is 4 and 2 (circle it if it is prime), 2 and 2;0832

whenever you have a pair of the same number, that comes out, so this is going to be 2.0846

Here is another 2 that is common; now, this one doesn't have a partner, doesn't have another 5 to come out of the radical.0855

So, only when they have a partner, only when there is two of the same number, can they come out.0870

2 comes out right there; then these 2's come out right there.0876

Now, the 5 has to stay in the radical; and then, you just multiply these two numbers.0881

And then, the square root of 25, we know, is 5; so this is going to be 4√5/5, and that is the distance.0888

Now, I know that this seems like a lot of work, but all we did was0897

construct a perpendicular segment by making the slope the negative reciprocal of this line.0904

Then, we found this point right here, where these two lines intersect; it has to be perpendicular.0915

So then, using this point and that point, we found the distance; and that is it--this is the distance between these two points.0923

All right, the next example: Find the distance between the two parallel lines.0938

OK, now, here I know that it has to be perpendicular wherever I want to find the distance,0943

as long as the segment that is made from the points between them has to be perpendicular.0958

Let's say I am going to use this point right here; then this is going to be perpendicular--that is going to be my distance.0965

Now, this point, I know is (0,2); this point, I know, is going to be 1 and 1/2...0978

1.5, or maybe 3/2 (the same thing): 3/2, and right here, that is going to be 1/2.0992

And I know that, because it is halfway between these two.0999

And the same here: the slope is 1, so everything is going to be half.1002

You can also, if you want, look at it this way: if I continue this on, then it is going to be halfway between this point right here1009

and this point right here, in the same way that this is halfway between this point and this point.1018

That is how I know that it is half.1026

This point is 3/2, 1, and 1/2; so now I have to find the distance between those two points.1028

The distance formula, again, is (x2 - x1), or (x1 - x2), squared, + (y2 - y1)2.1039

I find the square root again; I am going to subtract the x's: 0 - 3/2, squared, plus 2 - 1/2, squared.1057

Then, this is going to be (3/2)2; (3/2)2 is 9/4, plus...2 - 1/2 is 3/2, or 1 and 1/2,1073

and I know that because 2 becomes 4/2, minus 1/2 is 3/2; that squared is 9/4.1090

The square root of...there is already a common denominator, so that is 18/4...(make sure this goes all the way down),1104

which is √18/√4; for √18, you don't have to do the factor tree,1115

because you know that 9, a perfect square, is a factor of 18; it is going to be 3√2/2.1122

All you do is make sure that you just have the two points, whenever you try to find the distance between a point and a line or two parallel lines.1136

Then make sure that you have the two points, so that the segment that connects them is going to be perpendicular to the lines.1145

And then, you just use the distance formula; it is (0 - 3/2)2 + (2 - 1/2)2.1152

And then, that way, you get (3/2)2, which is 9/4, and then this, which is 3/2, squared, is going to be 9/4 again.1163

√18 over √4 simplifies to 3√2/2.1173

That is it for this lesson; thank you for watching Educator.com.1184