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Algebra 2 Midpoint and Distance Formulas

Name: Algebra 2: Midpoint and Distance Formulas
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Section 10: Conic Sections: Lecture 1 | 32:42 min

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	2 answers Last reply by: Matthew Deng Sat Jul 25, 2020 6:26 PM Post by julius mogyorossy on March 26, 2014 Forget what I said above, Dr. Carleen did not leave out a factor, 2, I apologize to Dr. Carleen, and to my fellow Educators. My head is really messed up right now, PTSD, but every now and then I get a sense of the God I shall be, it is so awesome, I wish I could record that state of mind on a DVD, and let you experience it.
	0 answers Post by julius mogyorossy on March 24, 2014 It seems Dr. Carleen left out a factor, 2, when finding the distance for the second example, in the first term of the first term, when she simplifies before squaring the two terms under the radical, but don't even take my word for it, I have bad PTSD, very stressed out, can't wait to be perfect soon and then take the Algebra Clep test. I so love, solving, writing equations, on my tablet.
	1 answer Last reply by: Dr Carleen Eaton Wed Nov 2, 2011 9:20 PM Post by Jonathan Taylor on October 30, 2011 1st example of distance formula 9+25 is not 36 it was 34 so the square root shold have been 34
	1 answer Last reply by: Dr Carleen Eaton Wed Nov 2, 2011 9:17 PM Post by Jonathan Taylor on October 25, 2011 I did not understand this question did u do the example as shown are did u make up a example
	2 answers Last reply by: Dr Carleen Eaton Thu Oct 13, 2011 9:05 PM Post by Manuela Fridman on October 6, 2011 I found 2 minor mistake in this video that I think should be corrected and reposted. The 1st one: in "example:distance" Dr. Eaton put the answer was 6. I believe that is incorrect because sq root 9+25= sq root 34 not 36 as Dr. Eaton put. Therefore the answer would be 5.83. The 2nd thing, i'm not sure if this makes a difference in the overall answer but in the "example 2:midpoint and" section Dr Eaton says to first put "y2" first but actually rights it using the "x2". It made it a little confusing, but of course i realized it was just a simple mistake. Thank you very much for all of your help though! Dr Eaton and the rest of the tutors are great! Just wish i was able to sign up for the year long membership without having to pay upfront for all the classes. $35 is a little high every month.

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Midpoint and Distance Formulas

Make sure that you know and understand how to use both of these formulas. They will be used in a lot of later work.
Remember that the order of the x coordinates in the distance formula does not matter. So take whichever difference is easier to compute. The same comment applies to the y coordinates.
After squaring the differences in the distance formula, be sure to take the positive square root of their sum.

Midpoint and Distance Formulas

Find the midpoint and distance of the segment with endpoints ( − 4, − 1),(2,4)

Use the midpoint formula M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) and distance formula D = √{(x₂ − x₁)² + (y₂ − y₁)²}
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [( − 4 + 2)/2],[( − 1 + 4)/2] ) = ( [( − 2)/2],[3/2] ) = ( − 1,[3/2] )

D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(2 − ( − 4))² + (4 − ( − 1))²} = √{(6)² + (5)²} = √{36 + 25} = √{61} ≈ 7.81

Find the midpoint and distance of the segment with endpoints ( − 3, − 5),(1,7)

Use the midpoint formula M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) and distance formula D = √{(x₂ − x₁)² + (y₂ − y₁)²}
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [( − 3 + 1)/2],[( − 5 + 7)/2] ) = ( [( − 2)/2],[2/2] ) = ( − 1,1 )

D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(1 − ( − 3))² + (7 − ( − 5))²} = √{(4)² + (12)²} = √{16 + 144} = √{160} ≈ 12.65

Find the midpoint and distance of the segment with endpoints (3, − 2),( − 2,5)

Use the midpoint formula M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) and distance formula D = √{(x₂ − x₁)² + (y₂ − y₁)²}
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [(3 − 2)/2],[( − 2 + 5)/2] ) = ( [1/2],[3/2] )

D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{( − 2 − (3))² + (5 − ( − 2))²} = √{( − 5)² + (7)²} = √{25 + 49} = √{74} ≈ 8.60

Find the midpoint and distance of the segment with endpoints ( − 3, − 5),(1,7)

Use the midpoint formula M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) and distance formula D = √{(x₂ − x₁)² + (y₂ − y₁)²}
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [( − 3 + 1)/2],[( − 5 + 7)/2] ) = ( [( − 2)/2],[2/2] ) = ( − 1,1 )

D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(1 − ( − 3))² + (7 − ( − 5))²} = √{(4)² + (12)²} = √{16 + 144} = √{160} ≈ 12.65

Triangle xyz has vertices x(5,9), y(1,1), and z(9,1). Find the length of the median from x to line yz.

Step 1 - Find the midpoint M between the points y and z.
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [(1 + 9)/2],[(1 + 1)/2] ) = ( 5,1 )
Step 2 - Find the Distance between the midpoint and point x.
D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(5 − (5))² + (1 − (9))²} = √{(0)² + ( − 8)²} = √{64} = 8

The length of the median from x to line yz is 8.

Triangle xyz has vertices x(8,8), y(2,4), and z(10,0). Find the length of the median from x to line yz.

Step 1 - Find the midpoint M between the points y and z.
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [(2 + 10)/2],[(4 + 0)/2] ) = ( 6,2 )
Step 2 - Find the Distance between the midpoint and point x.
D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(6 − (8))² + (2 − (8))²} = √{(2)² + ( − 6)²} = √{4 + 36} = √{40} ≈ 6.32

The length of the median from x to line yz is ≈ 6.32

Triangle xyz has vertices x( − 4, − 8), y(1,2), and z(9, − 4). Find the length of the median from x to line yz.

Step 1 - Find the midpoint M between the points y and z.
M = ( [(x₁ + x₂)/2],[(y₁ + y₂)/2] ) = ( [(1 + 9)/2],[(2 + − 4)/2] ) = ( 5, − 1 )
Step 2 - Find the Distance between the midpoint and point x.
D = √{(x₂ − x₁)² + (y₂ − y₁)²} = √{(5 − ( − 4))² + ( − 1 − ( − 8))²} = √{(9)² + (7)²} = √{81 + 49} = √{130} ≈ 11.40

The length of the median from x to line yz is ≈ 11.40

Find the perimeter and area of the triangle with vertices at A(2,4) B(5,2) C(6,10).

Step 1. Find lenths for sides ―AC ,―AB and―BC and add them to find the perimeter.
―AC : (2,4)(6,10)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(6 − (2))² + (10 − (4))²} = √{(4)² + (6)²} = √{16 + 36} = √{52} ≈ 7.21
―AB : (2,4)(5,2)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(5 − (2))² + (2 − (4))²} = √{(3)² + ( − 2)²} = √{9 + 4} = √{13} ≈ 3.61.
―BC : (5,2),(6,10)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(6 − (5))² + (10 − (2))²} = √{(1)² + (8)²} = √{1 + 64} = √{65} ≈ 8.06
Perimeter = ―AC +―AB +―BC = 7.21 + 3.61 + 8.06 = 18.88
Step 2: Find the area of the triangle using the formula A = [1/2]bh
For this formula to work, you must have a Right Triangle.
Recall that perpendicular lines form a 90 degree angle, and the product of their slopes is − 1.
Check the slopes of AB and AC to see if their product of their slopes equals − 1.
m_AB = [rise/run] =
m_AC = [rise/run] =
m_AB = [rise/run] = [( − 2)/3] = − [2/3]
m_AC = [rise/run] = [6/4] = [3/2]
Check the product of the slopes
m_AB*m_AC = [( − 2)/3]*[3/2] = − 1
Find the area

A = [1/2]bh = [1/2](3.61)(7.21) = 13

Find the perimeter and area of the triangle with vertices at A(2,4) B(4, - 2) C(11,7).

Step 1. Find lenths for sides ―AC ,―AB and―BC and add them to find the perimeter.
―AC : (2,4)(11,7)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(11 − (2))² + (7 − (4))²} = √{(9)² + (3)²} = √{81 + 9} = √{90} ≈ 9.49
―AB : (2,4)(4, − 2)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(4 − (2))² + ( − 2 − (4))²} = √{(2)² + ( − 6)²} = √{4 + 36} = √{40} ≈ 6.32
―BC : (4, − 2),(11,7)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(11 − (4))² + (7 − ( − 2))²} = √{(7)² + (9)²} = √{49 + 81} = √{130} ≈ 11.40
Perimeter = ―AC +―AB +―BC = 9.49 + 6.32 + 11.40 = 27.21
Step 2: Find the area of the triangle using the formula A = [1/2]bh
For this formula to work, you must have a Right Triangle.
Recall that perpendicular lines form a 90 degree angle, and the product of their slopes is − 1.
Check the slopes of AB and AC to see if their product of their slopes equals − 1.
m_AB = [rise/run] =
m_AC = [rise/run] =
m_AB = [rise/run] = [( − 6)/2] = − 3
m_AC = [rise/run] = [3/9] = [1/3]
Check the product of the slopes
m_AB*m_AC = − 3*[1/3] = − 1
Find the area

A = [1/2]bh = [1/2](6.32)(9.49) = 30

Find the perimeter and area of the triangle with vertices at A(1, - 1) B(2,2) C(7, - 3).

Step 1. Find lenths for sides ―AC ,―AB and ―BC and add them to find the perimeter.
―AB : (1, − 1)(2,2)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(2 − (1))² + (2 − ( − 1))²} = √{(1)² + (3)²} = √{1 + 9} = √{10} ≈ 3.16
―AC : (1, − 1)(7, − 3)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(7 − (1))² + ( − 3 − ( − 1))²} = √{(6)² + ( − 2)²} = √{36 + 4} = √{40} ≈ 6.32
―BC : (2,2),(7, − 3)
D = √{(x₂ − x₁)² + (y₂ − y₁)²} =
√{(7 − (2))² + ( − 3 − (2))²} = √{(5)² + ( − 5)²} = √{25 + 25} = √{50} ≈ 7.07
Perimeter = ―AC +―AB +―BC = 3.16 + 6.32 + 7.07 = 16.55
Step 2: Find the area of the triangle using the formula A = [1/2]bh
For this formula to work, you must have a Right Triangle.
Recall that perpendicular lines form a 90 degree angle, and the product of their slopes is − 1.
Check the slopes of AB and AC to see if their product of their slopes equals − 1.
m_AB = [rise/run] =
m_AC = [rise/run] =
m_AB = [rise/run] = [3/1] = 3
m_AC = [rise/run] = [( − 2)/6] = − [1/3]
Check the product of the slopes
m_AB*m_AC = 3*[( − 1)/3] = − 1
Find the area

A = [1/2]bh = [1/2](3.16)(6.32) = 10

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Answer

Midpoint and Distance Formulas

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

Midpoint Formula

Example: Midpoint

Distance Formula

Example: Distance

Example 1: Midpoint and Distance

Example 2: Midpoint and Distance

Example 3: Median Length

Example 4: Perimeter and Area

Intro 0:00
Midpoint Formula 0:15

Example: Midpoint

Distance Formula 2:30

Example: Distance

Example 1: Midpoint and Distance 4:58
Example 2: Midpoint and Distance 8:07
Example 3: Median Length 18:51
Example 4: Perimeter and Area 23:36

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Midpoint and Distance Formulas

Welcome to Educator.com.0000

Today is the first in a series of lectures on conic sections.0002

And we are going to start out with a review of two formulas that you will be applying later on, when we work with circles and hyperbolas and other conic sections.0005

Starting out with the midpoint formula: this formula gives you the midpoint of line segments with endpoints (x₁,y₁) and (x₂,y₂).0015

So, let's take an example, just to illustrate this: let's say you were asked to find the midpoint of a line segment with endpoints at (2,3) and (5,1).0028

If these were given as the endpoints, and you were asked to find the midpoint, you could go ahead and apply this formula.0044

So, to visualize this, let's draw out this line segment: this is at (2,3); that is one endpoint, and (5,1) is the other endpoint.0053

I draw a line between these two; and I am looking for the midpoint, which is somewhere around here.0065

Looking at this formula, it is actually somewhat intuitive, because what you are doing is finding0072

the average of the x-values and the average of the y-values, which will give you the middle of each (the midpoint).0078

So, x₁ + x₂ would give me 2 + 5 (the two x-values), divided by 2.0085

That is going to give me the x-coordinate of the midpoint of this line segment.0095

For the y-coordinate, I am going to add the two y's and divide by 2; so I will average those two y-values, which are 3 and 1.0099

2 + 5 gives me 7/2; 3 + 1 is 4/2; I can simplify to (7/2,2); and I could rewrite this, even, as (3 1/2,2) to make it a little easier to visualize on the graph.0110

This is the midpoint; it is going to be at 3 1/2 (that is going to be my x-coordinate); and 2 is going to be my y-coordinate.0128

And that is the midpoint; this is a fairly straightforward formula; you will need to apply it in a little while, when we start working with circles.0138

Also, to review the distance formula: if you recall, the distance formula is based on the Pythagorean theorem.0151

And the distance formula tells us that we can find the distance between two sets of points0159

if we take the square root of x₂ - x₁, squared, plus y₂ - y₁, squared.0163

Now recall: if I am given two points; if I am asked to find the distance between a set of two points, (2,1) and (5,6),0172

I can assign either one to be...I could assign this as (x₁,y₁), (x₂,y₂);0192

or I could do it the other way around: I could say this is (x₂,y₂); this is (x₁,y₁).0198

It doesn't matter, as long as you assign it and stick with that; you don't mix and match and say this is (x₁,y₂).0202

Just assign either set; it doesn't matter which set you call which; just stay consistent with the order that you use the points in.0209

OK, so let's look at what this would graph out to.0216

(2,1) and (5,6) would be right about up there; so I have a line segment, and I am asked to find the distance.0221

So, the distance is going to be equal to the square root of...I am going to call this (x₁,y₁) and this (x₂,y₂).0231

So, x₂ is 5, minus x₁, which is 2; take that squared,0239

and add it to y₂, which is 6, minus y₁, which is 1, squared.0247

Therefore, distance equals 3², plus 6 minus 1 is 5².0253

The distance equals the square root of 3², which is 9, plus 25.0259

Therefore, distance equals √36; distance equals 6.0269

The distance from this point to this point is equal to 6.0277

And again, this is review from an earlier lecture, when we discussed the Pythagorean theorem in Algebra I0282

and talked about how the distance formula comes from that.0288

So, you can always go back and review that information; but this is the application of the distance formula.0291

In the first example, we are asked to find the midpoint and distance of the segments with these endpoints (-9,-7) and (-3,-1).0299

So, recall that the midpoint formula just involves finding the average of the x-coordinates0310

of the two points, and the average of the y-coordinates of the two points.0315

Applying these values to this set of equations gives me -9 + -3, divided by 2, and -7 + -1, divided by 2.0324

Adding -9 and -3 gives me -12, divided by 2; and that is -8 divided by 2.0349

This becomes -6 (-12/2 is -6), and then -8/2 is -4.0359

So, that is the midpoint; that was the midpoint formula.0367

Recall that the distance formula is (x₂ - x₁)² + (y₂ - y₁)².0377

I am going to assign this as (x₁,y₁) and this as (x₂,y₂).0392

Again, it doesn't matter; you can do it the other way around.0404

Distance equals...I have x₂ as -3, minus -9, all of this squared; plus y₂, which is -1, minus -7, and that whole thing squared.0406

This gives me -3; a negative and a negative is a positive, so plus 9, squared, plus -1...and a negative and a negative gives me + 7...squared.0433

Therefore, distance equals the square root of...9 - 3 is 6, squared, and then 7 - 1 is also 6, squared.0444

Distance equals √(36 + 36), or radic;72.0453

But I can take this a step farther, and say, "OK, this is the square root of 36 times 2," and then simplify that,0461

because this is a perfect square, to 6√2.0467

So, I found the midpoint; it is (-6,-4); that is the midpoint of this segment with these endpoints.0475

And the distance between these endpoints is 6√2.0481

In Example 2, we are asked to find the midpoint and distance of the segment with these endpoints.0488

We go about it as we usually do; but this time we are working with radicals,0493

so we have to be really careful that we keep everything straight.0496

Recall that the midpoint formula is the average of the x-coordinates, and then the average of the y-coordinates.0499

Therefore, the midpoint equals x₁...the square root of 2 + 3, plus x₂, 2√2 - 4√3, divided by 2;0512

for the y-coordinate of the midpoint, we are going to get √3 - 5, plus 4√3, plus 2√5.0526

And then, we simplify this as much as we can.0536

Here, I have two like radicals: they have the same radicand, so I can combine those to make this 3√2.0540

I have a constant, and then this -4√3.0549

So, this is the x-coordinate of the midpoint; the y-coordinate is √3 here and 4√3; those can be combined into 5√3;0555

and then, let's see: this actually should have a radical over it, because that is a radical up there;0567

I have -√5 + 2√5; that is going to leave me with just + √5, divided by 2.0577

And the midpoint is given by this; it is a little bit messy-looking, but it is correct--you can't simplify, really, any farther.0585

So, we are just going to leave it as it is.0592

Now, we are also asked to find the distance; so I am going to go ahead and work that out here.0594

We found the midpoint; we are working with these same endpoints, but we are finding the distance.0599

Recall that the distance formula is the square root of (x₂ - x₁)² + (x₂ - y₁)².0603

So, I am going to let this be (x₁,y₁), (x₂,y₂).0617

I could have done it the other way around; it doesn't matter, as long as you are consistent.0625

Therefore, the distance is going to be equal to y₂, which is 2√2, minus 4√3;0630

and I am going to take that, and from that I am going to subtract y₁, which is √2 + 3.0645

So, this covers my (x₂ - x₁)².0657

I am going to add that to (y₂ - y₁): I go over here, and I have y₂; that is 4√3 + 2√5.0664

And I am going to subtract y₁ from that, which is over here; and that is √3 - √5; and this whole thing is also going to be squared.0678

Let's apply these negative signs to everything inside the parentheses, so we can start doing some combining.0692

This is going to give me 2√2 - 4√3; negative...that is going to give me -√2;0700

apply the negative to that 3--it is going to give me -3; all of this is going to be squared.0713

Plus 4√3, plus 2√5...apply the negative to each term inside the parentheses to get -√3;0718

and this negative, and then the negative √5, gives me a positive √5; and this whole thing is squared.0728

All right, so let's see if I can do some combining to simplify a little bit before I start squaring everything,0736

because that is going to be the most difficult step.0742

All right, this gives me 2√2, and this is minus √2; so I can combine these two to get √2.0747

-4√3, minus 3; this whole thing squared, plus 4√3 - √3--that simplifies to 3√3.0757

2√5 + √5 gives me 3√5; squared.0773

Now, I need to just square everything, and then combine it.0780

There is no easy way around this first one; what I am going to do is multiply √2 times itself, times this, times this,0784

and then go on with the second, and finally the third, term.0792

So, this √2 times √2 is simply going to give me 2; √2 times -4√3 is going to give me -4... 2 times 3 is √6.0795

Then, √2 times -3 is going to give me -3√2.0817

OK, now I take -4√3 and multiply it by this first term, √2, to get -4; 3 times 2 is 6.0828

When I multiply this times itself, I am going to get -4 times -4; that is going to give me 16,0840

times √3 times √3 is going to give me 3.0846

Then, I multiply this times -3 to get -4 times -3 is 12√3.0851

Finally, multiplying -3 times √2 gives me -3√2; -3 times -4 is 12√3; -3 times -3 is 12√3.0861

And then, -3 times -3 is 9; all of this is the trinomial squared; now let's square the binomial.0878

3√3 times 3√3 is going to be...3 times 3 is 9; and then √3 times √3 is just going to be 3.0887

So, looking up here, just to make this a little clearer: 3√3 + 3√5, times itself (squared)...0902

I am going to do my first terms, and I get this; then I do the outer terms, plus the inner terms,0919

which would just be this times this times 2; so that is going to give me 2 times 3 times 3, which is 9,0924

times 3 times 5, with a radical over it; so that is this.0941

So, what I did is took...really, it is just outer plus inner, which is going to give me the outer, which is 9√15, plus the inner, which also 9√15.0951

And this is going to end up giving me 18√15, which is the same as what I have here.0969

And then, finally, the last terms are going to give me 3 times 3 is 9, and then √5 times √5 is just going to give me 5.0976

OK, now combining what we can in order to just make this a bit simpler: I have a constant here;0989

I am just going to cluster all of my constants together in the beginning; that is 2; 16 times 3 is 48, so that is 48;1007

what other constants do I have?--I have a 9, and then I have 27, and I have 45.1016

All right, so those are my constants.1035

Now, for terms with a √6 in them, I have 2 of those: -4√6 and -4√6 is going to give me -8√6.1037

So, I took care of the constants and the terms with a radicand of 6.1048

Next, let's look at terms with a radicand of 2: -3√2, and I have -3√2 here; and that is it.1052

I add those two together, and I am going to get -6√2.1062

So, these are all taken care of: √2, √6, constant: now I have √3.1066

12√3--do I have any other terms like that?--yes: 12√3, so I have 2 of those; and that is going to be...12 and 12 is 24√3.1072

Let's see what else I have left: I took care of those, those...that just leaves me with...2 times 9 is 18 times √15.1083

Therefore, the distance equals...putting this all together, if you added these up, you will get 131 - 8√6 - 6√2 + 24 √3 + 18 √15.1094

So, this is the distance; and this was really a lot of practice just working with multiplying and adding and subtracting radicals.1115

We found the midpoint in the previous slide; and here is the distance for the segment with these endpoints, applying the distance formula.1123

Example 3: Triangle XYZ has vertices X (4,9), Y (8,-9), and Z (-5,2).1132

Find the length of the median from X to line YZ.1143

Definitely, a sketch would help us to solve this; so let's sketch this triangle.1146

And it has...we will call this X at (4,9); at (8,-9), we are going to have Y; and then, Z is going to be over here at (-5,2).1152

The median is going to be a line going here from here right to this midpoint.1173

So, what we are asked to find is the length: we can use the distance formula to find the length.1179

But in order to use the distance formula, I need this endpoint and this endpoint.1185

But I don't have this endpoint; however, I can find it, because if you look at this, this is going to draw out right to here, right in the middle.1190

So, this is the midpoint; if I find the midpoint of YZ, then I have the endpoint of this median.1197

Using the midpoint formula: (x₁ + x₂)/2, and then for the y-coordinate, (y₁ + y₂)/2.1208

So, I am looking for the midpoint of YZ: that is going to give me...1227

for Y, the x-value is 8; for Z, it is -5; divided by 2; for Y, the y-value is -9; for Z, the y-value is 2; divided by 2.1235

This is going to give me a midpoint of 8 - 5; that is 3/2; -9 + 2 is -7/2.1252

This is the midpoint, which is (3/2, -7/2); now I can use my distance formula,1262

distance equals √((x₂ - x ₁)² + (y₂ - y₁)²).1271

I can use that distance formula to find this.1282

And I am going to call (4,9)...I need the distance from x to this point m...I am going to call this1283

(x₁,y₁), and this (x₂,y₂), applying my distance formula.1292

So first, x₂, which is 3/2, minus x₁, which is 4--I am going to rewrite that as 8/2 to make my subtraction easier.1299

But this is just 4; squared; plus y₂, which is -7/2, minus 9, which I am going to rewrite as 18/2.1311

Again, I am just moving on to the next step of finding the common denominator.1324

But this is my (x₁,y₁), (x₂,y₂); that is where these came from.1328

Therefore, the distance equals...3/2 - 8/2 is -5/2; and we are going to square that;1339

plus -18 and -7 combines to -25/2, squared.1347

Therefore, the distance equals the square root of...this is 25 (5 squared), over 4, plus...25 squared is actually 625, divided by 4,1357

which equals the square root of 625 and 25 is 650, and they have the common denominator of 4.1370

And you could leave it like this, or take it a step farther and write this as 650 times 1/4, which equals 1/2.1379

And you could even look and see if there are any other perfect squares that you could factor out.1389

But this does give us the length of the median; in order to find the length of the median, we had to find this other endpoint,1393

which conveniently was the midpoint of YZ, so we found the midpoint of YZ and used that as the other endpoint.1401

Then I had X as an endpoint and the midpoint as the endpoint.1408

Plug that into the distance formula to get the length.1412

Working with triangles again: find the perimeter and area of the triangle with vertices at (1,-4), (-1,-2), and (6,1).1418

So, to help visualize this, we are going to draw it out on the coordinate plane.1427

(1,-4) is my first point; (-1,-2) is my second point; and then 6 is going to be about here; this is 6; 1 will be right here.1433

Let's see what this triangle looks like, and if the graph can help me.1449

Now, I am asked to find the perimeter and area: if I want to find the area of a right triangle,1459

I know that it is going to be 1/2 the base times height.1462

The problem is that I don't know if I am working with a right triangle; I may be; but this definitely does not look like a right angle.1465

This may be a right angle, but I am not sure.1472

In order to determine if it is a right angle (let's call this A, B, and C)--is this a right angle?--1475

well, if it is a right angle, that means that this AC and BC are going to be perpendicular.1485

And recall that perpendicular lines...the product of their slope equals -1.1492

So, what I am going to do is go ahead and find the slope of these two.1510

And I am going to determine what this slope is of AC; what BC is; find their product; and if they are a right angle, then I can proceed.1515

So, this point right here is (-1,-2); this point is (1,-4); and this point is (6,1).1523

Slope is just change in y, divided by change in x.1534

So, let's find the slope of AC: that is going to be the change in y (-2 - -4), divided by change in x (which is -1 - 1).1538

So, this is going to be -2 plus 4, divided by -2; this is going to give me 4 divided by -2, which is -2.1561

I just double-check this: -2 minus -4...actually, correction: 4 minus 2 is going to give me, of course, 2; and the slope, therefore, will be -1.1579

OK, now I also need to find, over here (to keep this separate), the slope of BC.1596

So, change in y: y is 1 - -4, over change in x: x is 6 - 1; this is going to be equal to 1 + 4, divided by 6 - 1 is 5, which equals 5/5.1606

So, the slope of BC equals 1.1630

Now, this means that, if I have the slope of BC, times the slope of AC, equals 1 times -1, or -1.1634

This is a right angle; I have a right triangle; I can use my formula, 1/2 the base times the height.1649

What I need to do next is find the length of these sides, using the distance formula.1656

So, let's start out by finding the length of side AC.1661

Recall your distance formula: for the distance formula, I am going to take x₂, side AC,1668

and I am going to call...it doesn't matter which one...I am going to call this, for this first one, (x₁,y₁), (x₂,y₂).1678

x₂ is -1, minus x₁, which is 1, squared, plus y₂, which is going to be -2, minus y₁, which is -4.1689

So, the length of AC equals...-1 and -1 is -2, squared; plus -2...and a negative and a negative is a positive, so -2 + 4 is going to give me 2, squared.1705

Therefore, the length of AC is going to equal the square root of 4 plus 4, which equals the square root of 8.1723

So, that is the length of AC.1733

Now, I need to find the length of another side to find the area.1735

I found AC; let's go for BC next--the length of BC.1739

I am going to make this (x₁,y₁),(x₂,y₂).1748

So, I am starting out with my x₂, which is 6, minus x₁, which is 1, squared,1753

plus x₂, which is 1, minus -4 (that is y₂ - y₁), squared,1761

equals the square root of...6 minus 1 is 5, squared; 1 minus -4...this becomes 1 + 4, so that is 5, squared;1773

So, this equals the square root of 25 + 25, or the square root of 50.1783

All right, so now I have two sides: I know that this side, AC, has a length of √8; and I know that BC is √50.1789

So, the area equals 1/2 the base times the height; so the area equals 1/2 (√8)(√50), which equals 1/2√400.1804

All right, and you could go on and then simplify this, because this would give you the perfect square of 20 times 20,1828

because 20² would give you 400; so then I could make this 1/2(20) is 10.1845

All right, now we still need to find the perimeter.1856

In order to find the perimeter, I need this third side; and I can use the Pythagorean theorem, because this is the hypotenuse.1859

And I know that a² + b² = c².1864

So, a² is...one side is the square root of 8, squared, plus b², the square root of 50 squared, equals c².1869

Well, the square root of 8 squared is 8, plus 50 (because the square root of 50 squared is going to be 50), equals c².1879

So, 58 = c²; therefore, c equals √58.1890

All right, so I found the area right here; now, the perimeter.1895

The perimeter equals the sum of the three sides: so that is √8 + √50 + √58.1907

And you could go on and do a little simplification.1920

You could pull the perfect squares out of here.1922

So, I could go a little farther with this and say, "OK, 8 is equal to 4 minus 2; 4 is a perfect square, so this is 2√2."1927

50 is the square root of 25 times 2, so that is going to give me 5√2, plus √58.1935

But you can't combine any of these, because they are not like radicals.1942

So, in this example, we applied the distance formula, as well as the Pythagorean theorem and some geometry,1947

to find the area and the perimeter of this triangle.1953

Thanks for visiting Educator.com; and that concludes this lesson on the midpoint and distance formulas.1957

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Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

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Midpoint and Distance Formulas

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro