Kamala, I apologize for not responding. I didn't see your question until today. Unfortunately, I don't know that solving that particular problem would help you more than solving the problem I did in the video, but I will say that you might find it easier if you first put both equations into slope-intercept form. Best regards! Shayne
l0ner, yes, I still check for comments and questions. While all of my students have graduated, others in the world are still finding the videos useful. All the best. Mr. Picard
this is the first video i've found explaining this topic that actually made sense to me! thank you for including a list of the steps, it was unbelievably helpful :D i've been trying to learn this for over two hours now, it is the only concept i've struggled with so far in geometry, and i think i finally get it! thank you so much, i can see from your thoughtful replies in the comments here that you truly care about helping students learn and you have no idea how much that means. you are a truly amazing teacher, keep doing what you're doing! the world needs more people like you :)
Astronymity: I'm really glad you found the video helpful. My videos, as with all other teachers' videos, vary in their clarity. It is always good to compare a number of sources and ask (1) which of these are consistent with the others? (2) which are consistent internally? (3) which are logically coherent? Every human source of information contains errors and biases, so careful as you go! All my best regards to you, Shayne
Thank you so much for helping me! I spent 3 1/2 hours working on my math homework, and you came in clutch with this informational video showing me how to do a hard problem. Thanks!!
This goes out to Blessings Suatia: I received your question via e-mail but could not locate it on my channel, so here is my reply. I assume you're asking about the perpendicular slope stated at 6:12 in the video. Perpendicular lines in the coordinate plane have slopes that are "opposite reciprocals." Think of a really large + sign in the coordinate plane and imagine it in any position except the position we normally see it in (that is, rotate it at least slightly). You'll find that whenever one of the bars has a positive slope, the other will necessarily have a negative slope. This is expressed mathematically by opposites (to find the opposite of a number, multiply by -1). You'll also find that, looking at a rotated +, when one bar has a steep slope, the other bar necessarily has a gentle slope. For example, if you rotate the + just slightly clockwise, so that the vertical bar has a slope of 4, the horizontal bar will have a slope of -1/4. 1/4 is the reciprocal of 4 (their product is 1), and -1/4 is the opposite reciprocal. So, in summary, to find the slope that is perpendicular to a given slope, determine the opposite reciprocal. For a good understanding of that concept, I suggest you go to mathbitsnotebook.com/Geometry/Equations/EQPerpendicularLines.html. Hope that helps!
Another easier, and quicker method I deduced(it's easy to come up with): Dist=(b2-b1)cos(arctan(m)) Where b2, b1, and m are the constants as in y=mx+b Tried it, it wroked. Anyhow, thank you for your demonstration. Very helpful.
Sharukh, I assume you are referring to 3:49 in the video? The distance formula is really not needed when finding the distance between horizontal or vertical lines (you can simply use the Ruler Postulate). To see the distance formula in action, scrub to 9:06. May all your studies be fruitful! Shayne
Being smart is more about how we pursue knowledge than about the amount we have. I believe you're smart because you are searching for knowledge. Keep it up.
This is way too easy and simple that what I have studied 😭😭tommorow I have EOT exam I hope I can do well! This is literally the best video on explaining the distance between two lines, thank you so much, very helpful you saved my life!
Excellent. This is the proof. As a teacher, it is easy to assume that your explanation is clear. But only when a student has understood is the explanation validated.
Dear Jae, This video has been up for almost 6 years and no one has asked your question before. It’s a bit more complicated than the subject of the video, but I’ll take a shot at answering you. Please note that there are a few ways to approach the problem. I have set out what I consider to be the easiest solution that a Geometry student should be able to understand, though it is possible that another math teach who is sharper than I would have a better answer for you. Either way, great job on asking this excellent question! Keep learning! First, I’ll run through a solution to a specific problem: Find the two lines that are parallel to and 5 units distant from the line y = 1/2x + 4 (note that in a plane, there will always be two parallel lines at any given distance; in three dimensions, there are an infinite number of such lines). I recommend you work this problem out on paper. Unfortunately, the comments section of RUclips does not allow for diagrams. To solve: 1. Identify a perpendicular line along which to work. In my opinion, the easiest line to choose is the perpendicular line that has the same y-intercept as the given line. So, in our example, I choose the line y=-2x + 4. You should already be able to “picture” or visualize the fact that we are looking for the two lines that are perpendicular to y = -2x + 4 at a distance of 5 units from y = 1/2x + 4. 2. Now, think of the distance, 5 units, as being the hypotenuse of a right triangle in which one leg is located along the y-axis and the other leg is parallel to the x-axis. We do not know the lengths of these legs, but we do know that their ratio is 2/1 (the absolute value of the slope of our perpendicular line). 3. Using the Pythagorean Theorem: y^2 + x^2 = 5^2 (where y is the length of the vertical leg along the y-axis and x is the length of the horizontal let parallel to the x-axis). Unfortunately, we will not be able to solve the Pythagorean Theorem so long as there are two variables. 4. Because we know the ratio of the two legs (or, in other words, we can define the length of one in terms of the other), we can eliminate one of the variables. In this case, y = 2x. Therefore, (2x)^2 + x^2 = 5^2. Simplifying: x^2 = 5. Further simplifying: x = +/- sqrt(5). Since y = 2x, y = +/-2sqrt(5). If you use your calculator, you’ll find that x = +/-2.24 and y = +/-4.47. 5. Now, it is IMPORTANT that you realize a couple things about these numbers. First, they are not coordinates, but rather the lengths of the legs of the two right triangles the hypotenuses of which lie between our y-intercept and the points that are on our parallel lines. Second, even though there are four potential pairs of legs [(2.24, 4.47), (2.24, -4.47), (-2.24, 4.47), and (-2.24, -4.47)], only two of them are correct. How do we know which two? By comparing the sign of the ratio of the legs to the sign of the slope of our perpendicular line. Here, our perpendicular line has slope -2. Therefore, we need to focus on the pairs (2.24, -4.47) and (-2.24, -4.47). 6. Now, we use these legs, or distances, to find a point on each of the two parallel lines. Our y-intercept is (0, 4). Therefore, our first point is found by adding 2.24 to the x-value and -4.47 to the y-value. Accordingly, our point on the first (lower) parallel line is (2.24, -0.47). Use the distance formula to make sure that is correct. To find the point on the second parallel line, add -2.24 to the x-value and 4.47 to the y-value of the y-intercept: (-2.24, 8.47). 7. Finally, use point-slope form (twice) to find the two equations (I’ll do just one in this example): y - (-0.47) = 1/2 (x - 2.24). Simplify: y + 0.47 = 1/2 x - 1.12; y = 1/2x - 1.59 - This is the first (lower) of the two parallel lines. Now, can we reduce this to a process? I’ll write out the steps and include one more example in parentheses. Find the two lines that are parallel to y = 3x - 2 and 3 units distant. 1. Identify the perpendicular slope. (-1/3) 2. Define one of the distances in terms of the other. In this case, x = 3y. 3. Use the Pythagorean Theorem to determine the vertical and horizontal distances: y^2 + (3y)^2 = 3^2; 10y^2 = 9; y^2 = 0.9; y = 0.95. 4. Since the horizontal distance is 3 times the vertical distance, the horizontal distance is 3 * 0.95, or 2.85. 5. Determine the two pairs of possible legs on the right triangles (remembering that the sign of their ratio will be the same as the sign of the perpendicular slope): (2.85, -0.95), (-2.85, 0.95) (Remember, even though I’ve written these as ordered pairs, they are pairs of distances, NOT coordinates). 6. Use the distances found in step 5 to determine the coordinates of the two points (one on each parallel line): (0, -2) + (2.85, -0.95) = (2.85, -2.95); (0, -2) + (-2.85, 0.95) = (-2.85, -1.05). 7. Use point-slope form to determine the equation of the first line: y-(-2.95) = 3(x-2.85); y + 2.95 = 3x - 8.55; y = 3x - 11.5 8. Use point-slope form to determine the equation of the second line: y-(-1.05) = 3(x - (-2.85); y + 1.05 = 3x + 8.55; y = 3x + 7.5 I hope this helps. You may need to run through it a few times in order to make sense of it.
This can be solved by using distance from a point to a line. Use the point (let's say y-intercept) from the other line and the equation of the other line.
Thank youuuu! I have a test this afternoon and it’s a retake cause I failed the first🥺 but this really helped! I think I’m gonna nail it. You deserve way more subscribers. Maybe you could do one on the distance between a curve and line🙏
Sambi, So glad to have helped you! I hope you did nail the test! Not sure I'll be able to post a video on finding the distance between a curve and a line, but I offer the following. First, without thinking about it to deeply, my first thought is that you would need to know some calculus in order to find the point on the curve at which slope of the curve is parallel to the given line. I assume (wrongly perhaps) that you watched this video and took your test in conjunction with a Geometry class and that you have not yet studies calculus. I wouldn't think a basic Geometry class provides the tools needed to solve this type of problem (though simplified version of the problem might be solved, for example where you are given the graph and can reasonably guess the point on the curve at which it comes nearest to the line). Second, once you determine the point on the line at which the slope is equal to the slope of the given line, you can then find the perpendicular line passing through that point. Then, solve a system of equations to find where that perpendicular line passes through the given line, and finally run the distance formula to determine the distance. Third, keep in mind that a system of equations which includes a line and a "curve" can have no real solutions or any number of real solutions, depending on the curve. If there are solutions, it may not make much sense to ask what the distance is between the curve and the line. Anyway, I digress. I hope you are loving math and mastering it, owning it. Best regards!
Wow I was wondering why he did all this things 😆 You can simply use the point (0,4) and the seconds line equation. Using this two informations, you can simply use the point to line distance formula 😶 and boom you are done! Anyways thank-you sir💯 I came to know many things!
Mihreteab Girma, Thank you for pointing out that there is a formula derived specifically for determining the distance between a point and a line. As it turns out, one can derive a single formula to distill any number of steps which are consistently taken in sequence. The question always is whether memorizing the new formula is worth the effort. Often it is; often it is not. In any event, if you would please set forth the point-to-line distance formula here, along with a clear explanation of its use, I imagine it might be of some utility to people. Please be certain not to express it concrete terms only (for example, above you say, "You can simply use the point (0,4) and the [second of the line equations]) as concrete examples, by themselves, cannot be used more broadly. Best regards, Shayne
Walter White, huh? And Lucifer? You have a name that is so much better than that (though he really was given a beautiful name); you should try to find it.
May everything you studied come back to you with clarity! May you have full command over every concept and immediate recall of every rule and definition.
I would have been glad if I had merely saved a few minutes of your life or helped you to improve your grade. Do remember, though, that the incalculable value of your life cannot be measured by a test, by a grade, or by any person's estimation of you. I do hope all goes well.
that is all good. Thank you for reminding me about point-slope form. and matching ys together. I usually go straight to slope-intercept from bc I know both things from beggining. My question is, what if your point B is not as perfect as (2,3) but both are weird decimals? Do I still have to plug it in the distance formula? What if it is on a test? IS there a way to get both points perfect or a shortcut?
El Castillo Imposible, First, you must leave behind the idea that there are "weird" decimals. That mindset is perfectly normal within many educational systems where the numbers you are asked to work with are nearly always integers, but that is not reality. Imagine if you were talking with several classmates, discussing your respective heights, and every one of you measured exactly four feet, five feet, or six feet tall. That would be strange indeed. Furthermore, assuming that you don't happen to be exactly four, five, or six feet tall, your doctor telling you that you are would be anything but perfect. No measurement is "perfectly" precise, so train yourself to be comfortable with decimals. Second, realize that every measurement is necessarily rounded to some fixed number of decimals. There are rules for determining how to do this, but they are beyond the scope of a typical Algebra or Geometry class, so your instructor (or text) typically tells you how to round (i.e. "Round your answer to the nearest tenth"). Third, yes, this process works no matter the form of the coordinates of B (integers, decimals, fractions). Fourth, yes, you must still use the distance formula to find the distance between A and B. Sixth, there is a shortcut which you have hinted at in your question. Because the equation of the perpendicular line, as identified in the video, will always have the same y-intercept and perpendicular slope as the first equation, and because you will always solve the system of equations (consisting of the second equation and the equation of the perpendicular line) by substitution, you can immediately substitute for y in the second equation the sum of the product of the perpendicular slope and x and the y-intercept of the first equation. For example: y = 1/2x + 2; y = 1/2x - 3 Therefore, -2x + 2 = 1/2x - 3. There is no need to graph these problems. I do it above to help students visualize the problem, but the Algebra is much quicker if you forego the visual representation. I hope this helps.
@@shaypic1 Thank you so much! Gracias. This helps so much! I will leave behind the mindset that decimals are weird. Really great video. The part where you show the graphing and algebraically solution. Thank you so much! Very helpful!
Glad to help. Beware the trap of thinking you can only learn if your teacher teaches. Everything you need to learn is within your reach. You need only search for it.
Deepak, you’re welcome. I hope it helped. And I apologize for the small writing; it’s sometimes hard to anticipate how the handwriting on the board will appear in a video. Best regards, Shayne
Hamza, I assume you’re referring to the slope found at ~6:07. The slope that is perpendicular to a given slope is always the “opposite reciprocal”. To find the opposite reciprocal of any slope, simply “flip” the fraction and multiply by -1. So the slope that is perpendicular to 2/3 is -3/2. A slope of -1/4 is perpendicular to a slope of 4 (note the denominator of 1 has not been written but can always be inferred). If you’re referring to something else, please reply with clarification.
Careful! Brief tangent: Is it peer or pear? I digress. Careful with assuming that information is ever for a particular grade. Information is for whatever student can appropriate it and be assisted by it. I did record the video for a 10th grade Geometry class, but if it helps, then it's for you (notice the use of that conditional).
If you're looking only for an example to follow, scrub to 10:36 (
You are the best teacher ever ❤️ thanks
Shayne Picard hellow sir can u teach me how to solve this qn.(Find the distance bet the parallel lines 2x-y+4=0 and 6x-3y=5)
Kamala, I apologize for not responding. I didn't see your question until today. Unfortunately, I don't know that solving that particular problem would help you more than solving the problem I did in the video, but I will say that you might find it easier if you first put both equations into slope-intercept form.
Best regards!
Shayne
This is one of the best video on the distance between parallel lines on RUclips. You deserve way more subscribers. Keep it up!
Akili KiKi, thanks for the kind words. I️ hope it helped bring clarity.
thank you mr. white. jessy is proud
I've been called a lot of things, but never Mr. White. Nonetheless, glad to help.
You're still active on this profile? Thanks. Serves us well.-
l0ner, yes, I still check for comments and questions. While all of my students have graduated, others in the world are still finding the videos useful. All the best. Mr. Picard
@@ghostinthemachine1760 Lmao.
@@particleonazock2246 Sup?
this is the first video i've found explaining this topic that actually made sense to me! thank you for including a list of the steps, it was unbelievably helpful :D i've been trying to learn this for over two hours now, it is the only concept i've struggled with so far in geometry, and i think i finally get it! thank you so much, i can see from your thoughtful replies in the comments here that you truly care about helping students learn and you have no idea how much that means. you are a truly amazing teacher, keep doing what you're doing! the world needs more people like you :)
Astronymity: I'm really glad you found the video helpful. My videos, as with all other teachers' videos, vary in their clarity. It is always good to compare a number of sources and ask (1) which of these are consistent with the others? (2) which are consistent internally? (3) which are logically coherent? Every human source of information contains errors and biases, so careful as you go! All my best regards to you,
Shayne
Thank you so much for helping me! I spent 3 1/2 hours working on my math homework, and you came in clutch with this informational video showing me how to do a hard problem. Thanks!!
God Bless you, you made me understand something that my teacher couldn't make me understand, I am forever grateful and ready for tomorrow's test.
Ana, so glad it helped. Feel free to reach out if there are other lessons you’re having trouble with.
Ana did you pass the test
This goes out to Blessings Suatia: I received your question via e-mail but could not locate it on my channel, so here is my reply. I assume you're asking about the perpendicular slope stated at 6:12 in the video.
Perpendicular lines in the coordinate plane have slopes that are "opposite reciprocals." Think of a really large + sign in the coordinate plane and imagine it in any position except the position we normally see it in (that is, rotate it at least slightly). You'll find that whenever one of the bars has a positive slope, the other will necessarily have a negative slope. This is expressed mathematically by opposites (to find the opposite of a number, multiply by -1).
You'll also find that, looking at a rotated +, when one bar has a steep slope, the other bar necessarily has a gentle slope. For example, if you rotate the + just slightly clockwise, so that the vertical bar has a slope of 4, the horizontal bar will have a slope of -1/4. 1/4 is the reciprocal of 4 (their product is 1), and -1/4 is the opposite reciprocal.
So, in summary, to find the slope that is perpendicular to a given slope, determine the opposite reciprocal. For a good understanding of that concept, I suggest you go to mathbitsnotebook.com/Geometry/Equations/EQPerpendicularLines.html. Hope that helps!
This was great. I followed what you did and got the correct answer and it's not so complicated anymore! Thank you!
Glad to help! Thank you for the positive feedback.
BEAUTIFUL JUST BEAUTIFUL .Thank you so much
Wow. You are so welcome. Thank you for the kind words.
Another easier, and quicker method I deduced(it's easy to come up with):
Dist=(b2-b1)cos(arctan(m))
Where b2, b1, and m are the constants as in y=mx+b
Tried it, it wroked.
Anyhow, thank you for your demonstration. Very helpful.
took me a day (learned trig recently) but i reached here as well
I could not figure this out for the life of me, thank you so much this made total sense!
Glad to help, Logan!
Skyler is proud of you mr white
Thanks...I think.
great 🤩🤩
Finding distance without any formula wow man exquisite!!
Sharukh,
I assume you are referring to 3:49 in the video? The distance formula is really not needed when finding the distance between horizontal or vertical lines (you can simply use the Ruler Postulate). To see the distance formula in action, scrub to 9:06.
May all your studies be fruitful!
Shayne
i feel smart now. thanks
Being smart is more about how we pursue knowledge than about the amount we have. I believe you're smart because you are searching for knowledge. Keep it up.
This is way too easy and simple that what I have studied 😭😭tommorow I have EOT exam I hope I can do well! This is literally the best video on explaining the distance between two lines, thank you so much, very helpful you saved my life!
Sorry for the late response, but I hope you aced that exam!
Your explanation is great... Thank you sir... I understand all concepts in the video... 👌👌👌👌
Excellent. This is the proof. As a teacher, it is easy to assume that your explanation is clear. But only when a student has understood is the explanation validated.
what if the distance is given and i have to find the other line?
Dear Jae,
This video has been up for almost 6 years and no one has asked your question before. It’s a bit more complicated than the subject of the video, but I’ll take a shot at answering you. Please note that there are a few ways to approach the problem. I have set out what I consider to be the easiest solution that a Geometry student should be able to understand, though it is possible that another math teach who is sharper than I would have a better answer for you. Either way, great job on asking this excellent question! Keep learning!
First, I’ll run through a solution to a specific problem:
Find the two lines that are parallel to and 5 units distant from the line y = 1/2x + 4 (note that in a plane, there will always be two parallel lines at any given distance; in three dimensions, there are an infinite number of such lines).
I recommend you work this problem out on paper. Unfortunately, the comments section of RUclips does not allow for diagrams.
To solve:
1. Identify a perpendicular line along which to work. In my opinion, the easiest line to choose is the perpendicular line that has the same y-intercept as the given line. So, in our example, I choose the line y=-2x + 4. You should already be able to “picture” or visualize the fact that we are looking for the two lines that are perpendicular to y = -2x + 4 at a distance of 5 units from y = 1/2x + 4.
2. Now, think of the distance, 5 units, as being the hypotenuse of a right triangle in which one leg is located along the y-axis and the other leg is parallel to the x-axis. We do not know the lengths of these legs, but we do know that their ratio is 2/1 (the absolute value of the slope of our perpendicular line).
3. Using the Pythagorean Theorem: y^2 + x^2 = 5^2 (where y is the length of the vertical leg along the y-axis and x is the length of the horizontal let parallel to the x-axis). Unfortunately, we will not be able to solve the Pythagorean Theorem so long as there are two variables.
4. Because we know the ratio of the two legs (or, in other words, we can define the length of one in terms of the other), we can eliminate one of the variables. In this case, y = 2x. Therefore, (2x)^2 + x^2 = 5^2. Simplifying: x^2 = 5. Further simplifying: x = +/- sqrt(5). Since y = 2x, y = +/-2sqrt(5). If you use your calculator, you’ll find that x = +/-2.24 and y = +/-4.47.
5. Now, it is IMPORTANT that you realize a couple things about these numbers. First, they are not coordinates, but rather the lengths of the legs of the two right triangles the hypotenuses of which lie between our y-intercept and the points that are on our parallel lines. Second, even though there are four potential pairs of legs [(2.24, 4.47), (2.24, -4.47), (-2.24, 4.47), and (-2.24, -4.47)], only two of them are correct. How do we know which two? By comparing the sign of the ratio of the legs to the sign of the slope of our perpendicular line. Here, our perpendicular line has slope -2. Therefore, we need to focus on the pairs (2.24, -4.47) and (-2.24, -4.47).
6. Now, we use these legs, or distances, to find a point on each of the two parallel lines. Our y-intercept is (0, 4). Therefore, our first point is found by adding 2.24 to the x-value and -4.47 to the y-value. Accordingly, our point on the first (lower) parallel line is (2.24, -0.47). Use the distance formula to make sure that is correct. To find the point on the second parallel line, add -2.24 to the x-value and 4.47 to the y-value of the y-intercept: (-2.24, 8.47).
7. Finally, use point-slope form (twice) to find the two equations (I’ll do just one in this example): y - (-0.47) = 1/2 (x - 2.24). Simplify: y + 0.47 = 1/2 x - 1.12; y = 1/2x - 1.59 - This is the first (lower) of the two parallel lines.
Now, can we reduce this to a process? I’ll write out the steps and include one more example in parentheses.
Find the two lines that are parallel to y = 3x - 2 and 3 units distant.
1. Identify the perpendicular slope. (-1/3)
2. Define one of the distances in terms of the other. In this case, x = 3y.
3. Use the Pythagorean Theorem to determine the vertical and horizontal distances: y^2 + (3y)^2 = 3^2; 10y^2 = 9; y^2 = 0.9; y = 0.95.
4. Since the horizontal distance is 3 times the vertical distance, the horizontal distance is 3 * 0.95, or 2.85.
5. Determine the two pairs of possible legs on the right triangles (remembering that the sign of their ratio will be the same as the sign of the perpendicular slope): (2.85, -0.95), (-2.85, 0.95) (Remember, even though I’ve written these as ordered pairs, they are pairs of distances, NOT coordinates).
6. Use the distances found in step 5 to determine the coordinates of the two points (one on each parallel line): (0, -2) + (2.85, -0.95) = (2.85, -2.95); (0, -2) + (-2.85, 0.95) = (-2.85, -1.05).
7. Use point-slope form to determine the equation of the first line: y-(-2.95) = 3(x-2.85); y + 2.95 = 3x - 8.55; y = 3x - 11.5
8. Use point-slope form to determine the equation of the second line: y-(-1.05) = 3(x - (-2.85); y + 1.05 = 3x + 8.55; y = 3x + 7.5
I hope this helps. You may need to run through it a few times in order to make sense of it.
Esta fue la clase mas comprensiva que eh visto en toda mi vida
Gracias profe
Que bueno que puedes entenderlo bien en ingles. Gracias por tus amables palabras.
This can be solved by using distance from a point to a line. Use the point (let's say y-intercept) from the other line and the equation of the other line.
True.
Thank youuuu! I have a test this afternoon and it’s a retake cause I failed the first🥺 but this really helped! I think I’m gonna nail it. You deserve way more subscribers. Maybe you could do one on the distance between a curve and line🙏
Sambi,
So glad to have helped you! I hope you did nail the test!
Not sure I'll be able to post a video on finding the distance between a curve and a line, but I offer the following. First, without thinking about it to deeply, my first thought is that you would need to know some calculus in order to find the point on the curve at which slope of the curve is parallel to the given line. I assume (wrongly perhaps) that you watched this video and took your test in conjunction with a Geometry class and that you have not yet studies calculus. I wouldn't think a basic Geometry class provides the tools needed to solve this type of problem (though simplified version of the problem might be solved, for example where you are given the graph and can reasonably guess the point on the curve at which it comes nearest to the line).
Second, once you determine the point on the line at which the slope is equal to the slope of the given line, you can then find the perpendicular line passing through that point. Then, solve a system of equations to find where that perpendicular line passes through the given line, and finally run the distance formula to determine the distance.
Third, keep in mind that a system of equations which includes a line and a "curve" can have no real solutions or any number of real solutions, depending on the curve. If there are solutions, it may not make much sense to ask what the distance is between the curve and the line.
Anyway, I digress. I hope you are loving math and mastering it, owning it. Best regards!
Teachers actually teach!! I’m jealous
Students actually study!! I'm grateful.
Thank you for this one.
You're welcome Shereece!
Mr white but for mathematics
Wow I was wondering why he did all this things 😆 You can simply use the point (0,4) and the seconds line equation. Using this two informations, you can simply use the point to line distance formula 😶 and boom you are done! Anyways thank-you sir💯 I came to know many things!
Mihreteab Girma,
Thank you for pointing out that there is a formula derived specifically for determining the distance between a point and a line. As it turns out, one can derive a single formula to distill any number of steps which are consistently taken in sequence. The question always is whether memorizing the new formula is worth the effort. Often it is; often it is not.
In any event, if you would please set forth the point-to-line distance formula here, along with a clear explanation of its use, I imagine it might be of some utility to people. Please be certain not to express it concrete terms only (for example, above you say, "You can simply use the point (0,4) and the [second of the line equations]) as concrete examples, by themselves, cannot be used more broadly.
Best regards,
Shayne
Lots of love from Nepal ❤️ I wish u were my math teacher ❤️
What a tremendous compliment. Thank you. शुभेक्षा सहित
Ke chha?
Thank u sirI have been searching for this ,and now I saw it .
You're welcome Heron. Glad I could help.
Thanks for the help Walter White of maths
Walter White, huh? And Lucifer? You have a name that is so much better than that (though he really was given a beautiful name); you should try to find it.
Ty so much
You're welcome. So much.
Thank you so, so much. I think I have a good chance at getting an A on tomorrow's test.
May everything you studied come back to you with clarity! May you have full command over every concept and immediate recall of every rule and definition.
@@shaypic1 Thank you :). I'm gonna rewatch your video a few times to really understand everything
u saved my life omg
I would have been glad if I had merely saved a few minutes of your life or helped you to improve your grade. Do remember, though, that the incalculable value of your life cannot be measured by a test, by a grade, or by any person's estimation of you. I do hope all goes well.
You look like walt from breaking bad
you should have written the words in big size..
wow. very helpful. thank you, and please start making videos again
Kush, glad it helped. I'll give some thought to your request, but I'm in a different line of work these days.
@@shaypic1 ok. Thank you
that is all good. Thank you for reminding me about point-slope form. and matching ys together. I usually go straight to slope-intercept from bc I know both things from beggining. My question is, what if your point B is not as perfect as (2,3) but both are weird decimals? Do I still have to plug it in the distance formula? What if it is on a test? IS there a way to get both points perfect or a shortcut?
El Castillo Imposible,
First, you must leave behind the idea that there are "weird" decimals. That mindset is perfectly normal within many educational systems where the numbers you are asked to work with are nearly always integers, but that is not reality. Imagine if you were talking with several classmates, discussing your respective heights, and every one of you measured exactly four feet, five feet, or six feet tall. That would be strange indeed. Furthermore, assuming that you don't happen to be exactly four, five, or six feet tall, your doctor telling you that you are would be anything but perfect. No measurement is "perfectly" precise, so train yourself to be comfortable with decimals.
Second, realize that every measurement is necessarily rounded to some fixed number of decimals. There are rules for determining how to do this, but they are beyond the scope of a typical Algebra or Geometry class, so your instructor (or text) typically tells you how to round (i.e. "Round your answer to the nearest tenth").
Third, yes, this process works no matter the form of the coordinates of B (integers, decimals, fractions).
Fourth, yes, you must still use the distance formula to find the distance between A and B.
Sixth, there is a shortcut which you have hinted at in your question. Because the equation of the perpendicular line, as identified in the video, will always have the same y-intercept and perpendicular slope as the first equation, and because you will always solve the system of equations (consisting of the second equation and the equation of the perpendicular line) by substitution, you can immediately substitute for y in the second equation the sum of the product of the perpendicular slope and x and the y-intercept of the first equation. For example:
y = 1/2x + 2; y = 1/2x - 3
Therefore, -2x + 2 = 1/2x - 3.
There is no need to graph these problems. I do it above to help students visualize the problem, but the Algebra is much quicker if you forego the visual representation.
I hope this helps.
@@shaypic1 Thank you so much! Gracias. This helps so much! I will leave behind the mindset that decimals are weird. Really great video. The part where you show the graphing and algebraically solution. Thank you so much! Very helpful!
Thank you sm. I have a quiz tomorrow and my teacher hasnt posted answers and we didnt have class on friday and its sunday now
Glad to help. Beware the trap of thinking you can only learn if your teacher teaches. Everything you need to learn is within your reach. You need only search for it.
sir, thank you very much. going to suscribe you
Deepak, you’re welcome. I hope it helped. And I apologize for the small writing; it’s sometimes hard to anticipate how the handwriting on the board will appear in a video. Best regards, Shayne
thanks
You’re quite welcome. I hope it helped.
thank you so much
You're welcome!
Thanks!
You bet!
How do you find the slope ? - 1/2
Hamza, I assume you’re referring to the slope found at ~6:07. The slope that is perpendicular to a given slope is always the “opposite reciprocal”. To find the opposite reciprocal of any slope, simply “flip” the fraction and multiply by -1. So the slope that is perpendicular to 2/3 is -3/2. A slope of -1/4 is perpendicular to a slope of 4 (note the denominator of 1 has not been written but can always be inferred).
If you’re referring to something else, please reply with clarification.
Thank you soo much again sir!
Master Cool, it's my pleasure, so long as the material helped you in some way. Best regards to you!
A like for you sir
Thank you!
Is this for grade 10??
Careful! Brief tangent: Is it peer or pear? I digress. Careful with assuming that information is ever for a particular grade. Information is for whatever student can appropriate it and be assisted by it. I did record the video for a 10th grade Geometry class, but if it helps, then it's for you (notice the use of that conditional).
Thank you sir
You're welcome. I hope the video was helpful.
jason satham
P😊please
come check me out john warns
Starts at 10:00
bakwas
कृपया बताएं कि यह कचरा क्यों है, यदि आप सक्षम हैं
This video is really good for understanding but some people aren't simply interested...