Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

Use the distance formula
Distance formula?
LOL

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question