### 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

―AB represents the distance.

―AB represents the distance.

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.

―AB represents the distance.

- Find point A on line p
- Draw ―AB ⊥ line q.

―AB represents the distance.

- ―AB represents the distance between line m and line n.

^{2}+ (9 − 7)

^{2}} = √{9 + 4} = √{13} .

―AB represents the distance.

- 2x + 4 = (x + 4) + 3

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}}

―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.

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
- Distance Between a Points and Line
- Distance Between Parallel Lines
- Extra Example 1: Drawing a Segment to Represent Distance
- Extra Example 2: Drawing a Segment to Represent Distance
- Extra Example 3: Graph, Plot, and Construct a Perpendicular Segment
- Extra Example 4: Distance Between Two Parallel Lines

- 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

### Geometry Online Course

### 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 distance*0067

*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 that*0407

*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

*That is your perpendicular segment.*0492

*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 (x _{2} - x_{1})^{2} + (y_{2} - y_{1})^{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 was*0897

*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 here*1009

*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 (x _{2} - x_{1}), or (x_{1} - x_{2}), squared, + (y_{2} - y_{1})^{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

0 answers

Post by Sai Nettyam on October 11, 2011

Ms. 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.