Question

What is the length of the longest side of a triangle that has vertices at (-5, 2) (1,-6) and (1,2)?

Answer 1

Use the distance formula

Answer 2

Distance formula?

Answer 3

LOL

Answer 4

ok its \[d=\sqrt{(x _{1}-x _{2})^2+(y _{1}-y _{2})^2} \]

Answer 5

I would also suggest labeling your coordinates

Answer 6

MathLegend ur not a math legend at all

Answer 7

A= (-5, 2)
B = (1,-6)
C = (1,2)

Answer 8

So when using the distance formula... you can solve for side AB first.

Answer 9

Thank you.

Answer 10

So using that formula, can you label the coordinates as (x1, y1) & (x2,y2)

Answer 11

A= (-5, 2)
B = (1,-6)

So lets solve for side AB first... so, "A" comes first right? So let that be your x1 & y1
Then, "B" can be your x2 & y2

Answer 12

Do you understand so far @ValentinaT ?

Answer 13

All I did was label them so that we can plug it into that distance formula.

Answer 14

@ValentinaT let me know when you get back so we can work this out together. :)

Answer 15

Yeah, I'm getting it, thank you.

Answer 16

\[\frac{ -6 - 2 }{ -5 - 1 } = \frac{ -8 }{ -6}\]

Answer 17

So for AB

(1+5)^2+(-6-2)^2

Answer 18

I'm sorry the formula is actually x2-x1 and y2-y1 the above poster just mixed up the two.

Answer 19

Okay.

Answer 20

(1+5)^2+(-6-2)^2
(6)^2+(-8)^2
36+64

Answer 21

36+64 = 100

Answer 22

\[\sqrt{100}\]

Answer 23

10

Answer 24

Now, we need the square root... because if you notice that entire formula had the square root symbol over it

Answer 25

Good.

Answer 26

So side AB = 10

Answer 27

So that is one side. So lets go for side BC

Answer 28

B = (1,-6)
x1 y1

C = (1,2)
x2 y2

Answer 29

@ValentinaT do you feel comfortable trying it out on your own? Tell me what you get and I'll check to see if your answer for side BC is correct.

Answer 30

\[\frac{ 2 - -6 }{ 1 - 1 } = \frac{ 8 }{ 0 }\]

Answer 31

Remember we are not trying to find a slope.

Answer 32

We are looking for the distance between the vertices.

Answer 33

(1-1)^2 + (2+6)^2

Answer 34

To get that all I did was plug it into the formula.

(x2-x1)^2+(y2-y1)^2

Answer 35

(1-1)^2 + (2+6)^2

Try solving that.

Answer 36

Okay, 0 + 64?

Answer 37

Good

Answer 38

0 + 64 = 64

\[\sqrt{64}\]

Answer 39

So, 8 as the square root.

Answer 40

Yes, so side BC = 8

Answer 41

Now, try side AC

Answer 42

(x2-x1)^2+(y2-y1)^2

Answer 43

A = -5, 2 = x1, y1
C = 1, 2 = x2, y2

\[\frac{ 1 - -5^2}{ 2 - 2^2 } = \frac{ 6 }{ 1 } \] so 36?

Answer 44

(x2-x1)^2+(y2-y1)^2

(1+5)^2 + (2-2)^2 Do you understand this step?

Answer 45

36 + 0 Yeah I get it, I just couldn't figure out how to write it.

Answer 46

Good so if you took the square root of 36

\[\sqrt{36}\]

Answer 47

So now, we know the length of the longest side is AB

Answer 48

Okay, thank you!