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

- anonymous

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

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- anonymous

Use the distance formula

- anonymous

Distance formula?

- anonymous

LOL

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## More answers

- anonymous

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

- MathLegend

I would also suggest labeling your coordinates

- anonymous

MathLegend ur not a math legend at all

- MathLegend

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

- MathLegend

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

- anonymous

Thank you.

- MathLegend

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

- hba

|dw:1355599903259:dw|

- MathLegend

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

- MathLegend

Do you understand so far @ValentinaT ?

- MathLegend

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

- MathLegend

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

- anonymous

Yeah, I'm getting it, thank you.

- anonymous

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

- MathLegend

So for AB
(1+5)^2+(-6-2)^2

- MathLegend

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

- anonymous

Okay.

- MathLegend

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

- MathLegend

36+64 = 100

- MathLegend

\[\sqrt{100}\]

- anonymous

10

- MathLegend

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

- MathLegend

Good.

- MathLegend

So side AB = 10

- MathLegend

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

- MathLegend

B = (1,-6)
x1 y1
C = (1,2)
x2 y2

- MathLegend

@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.

- anonymous

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

- MathLegend

Remember we are not trying to find a slope.

- MathLegend

We are looking for the distance between the vertices.

- MathLegend

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

- MathLegend

To get that all I did was plug it into the formula.
(x2-x1)^2+(y2-y1)^2

- MathLegend

(1-1)^2 + (2+6)^2
Try solving that.

- anonymous

Okay, 0 + 64?

- MathLegend

Good

- MathLegend

0 + 64 = 64
\[\sqrt{64}\]

- anonymous

So, 8 as the square root.

- MathLegend

Yes, so side BC = 8

- MathLegend

Now, try side AC

- MathLegend

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

- anonymous

A = -5, 2 = x1, y1
C = 1, 2 = x2, y2
\[\frac{ 1 - -5^2}{ 2 - 2^2 } = \frac{ 6 }{ 1 } \] so 36?

- MathLegend

(x2-x1)^2+(y2-y1)^2
(1+5)^2 + (2-2)^2 Do you understand this step?

- anonymous

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

- MathLegend

Good so if you took the square root of 36
\[\sqrt{36}\]

- MathLegend

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

- anonymous

Okay, thank you!

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