anonymous
  • anonymous
What is the length of the longest side of a triangle that has vertices at (-5, 2) (1,-6) and (1,2)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Use the distance formula
anonymous
  • anonymous
Distance formula?
anonymous
  • anonymous
LOL

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More answers

anonymous
  • anonymous
ok its \[d=\sqrt{(x _{1}-x _{2})^2+(y _{1}-y _{2})^2} \]
MathLegend
  • MathLegend
I would also suggest labeling your coordinates
anonymous
  • anonymous
MathLegend ur not a math legend at all
MathLegend
  • MathLegend
A= (-5, 2) B = (1,-6) C = (1,2)
MathLegend
  • MathLegend
So when using the distance formula... you can solve for side AB first.
anonymous
  • anonymous
Thank you.
MathLegend
  • MathLegend
So using that formula, can you label the coordinates as (x1, y1) & (x2,y2)
hba
  • hba
|dw:1355599903259:dw|
MathLegend
  • 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
  • MathLegend
Do you understand so far @ValentinaT ?
MathLegend
  • MathLegend
All I did was label them so that we can plug it into that distance formula.
MathLegend
  • MathLegend
@ValentinaT let me know when you get back so we can work this out together. :)
anonymous
  • anonymous
Yeah, I'm getting it, thank you.
anonymous
  • anonymous
\[\frac{ -6 - 2 }{ -5 - 1 } = \frac{ -8 }{ -6}\]
MathLegend
  • MathLegend
So for AB (1+5)^2+(-6-2)^2
MathLegend
  • MathLegend
I'm sorry the formula is actually x2-x1 and y2-y1 the above poster just mixed up the two.
anonymous
  • anonymous
Okay.
MathLegend
  • MathLegend
(1+5)^2+(-6-2)^2 (6)^2+(-8)^2 36+64
MathLegend
  • MathLegend
36+64 = 100
MathLegend
  • MathLegend
\[\sqrt{100}\]
anonymous
  • anonymous
10
MathLegend
  • MathLegend
Now, we need the square root... because if you notice that entire formula had the square root symbol over it
MathLegend
  • MathLegend
Good.
MathLegend
  • MathLegend
So side AB = 10
MathLegend
  • MathLegend
So that is one side. So lets go for side BC
MathLegend
  • MathLegend
B = (1,-6) x1 y1 C = (1,2) x2 y2
MathLegend
  • 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
  • anonymous
\[\frac{ 2 - -6 }{ 1 - 1 } = \frac{ 8 }{ 0 }\]
MathLegend
  • MathLegend
Remember we are not trying to find a slope.
MathLegend
  • MathLegend
We are looking for the distance between the vertices.
MathLegend
  • MathLegend
(1-1)^2 + (2+6)^2
MathLegend
  • MathLegend
To get that all I did was plug it into the formula. (x2-x1)^2+(y2-y1)^2
MathLegend
  • MathLegend
(1-1)^2 + (2+6)^2 Try solving that.
anonymous
  • anonymous
Okay, 0 + 64?
MathLegend
  • MathLegend
Good
MathLegend
  • MathLegend
0 + 64 = 64 \[\sqrt{64}\]
anonymous
  • anonymous
So, 8 as the square root.
MathLegend
  • MathLegend
Yes, so side BC = 8
MathLegend
  • MathLegend
Now, try side AC
MathLegend
  • MathLegend
(x2-x1)^2+(y2-y1)^2
anonymous
  • anonymous
A = -5, 2 = x1, y1 C = 1, 2 = x2, y2 \[\frac{ 1 - -5^2}{ 2 - 2^2 } = \frac{ 6 }{ 1 } \] so 36?
MathLegend
  • MathLegend
(x2-x1)^2+(y2-y1)^2 (1+5)^2 + (2-2)^2 Do you understand this step?
anonymous
  • anonymous
36 + 0 Yeah I get it, I just couldn't figure out how to write it.
MathLegend
  • MathLegend
Good so if you took the square root of 36 \[\sqrt{36}\]
MathLegend
  • MathLegend
So now, we know the length of the longest side is AB
anonymous
  • anonymous
Okay, thank you!

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