Thread: Hi, quick distance between points ?

1. Hi, quick distance between points ?

Hello everyone! I can't believe I actually found a forum for math! Mathletes unite, Well it's my last class to finish up my degree and I can't believe I waited till last minute to take it! now its beating me up!

I have a few questions, but will post one at a time in this thread so I can work them out to to understand them, thank you. The first one is

**Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint. (Hint: let (x,y) be the unknown endpoint. Apply the midpoint formula, and solve the 2 equations for x and y.)

Midpoint (27,1), endpoint (19,-1)

thank you so much its due before midnight!!

2. $\left(\frac{x+19}{2} \, , \frac{y + (-1)}{2} \right) = (27,1)$

solve for x and y.

3. I dont know how?

4. Originally Posted by Dillon658
I dont know how?

$\left(\frac{x+19}{2} \, , \frac{y -1}{2} \right) = (27,1)$

Compare the x and y coordinates.

$\frac{x+19}{2}=27\implies x=......$ and $\frac{y-1}{2}=1\implies y=......$

This shouldn't be much of a problem now. I'm sure you can take it from here.

--Chris

5. I guess im looking for the answer....

Theres so many different ways that I have in my head right now it all looks japanese...

6. Originally Posted by Dillon658
I guess im looking for the answer....

Theres so many different ways that I have in my head right now it all looks japanese...
I'll solve for y, and leave it for you to solve for x:

$\frac{y-1}{2}=1$

multiply both sides by 2 to get $y-1=2$

To isolate y, add 1 to both sides. We should now see that $\color{red}\boxed{y=3}$

Does this make sense? Try to find x in a similar way.

--Chris

7. is it 11?

8. Originally Posted by Dillon658
is it 11?
No. Did you first multiply both sides by 2? You should get $x+19=54\implies x=......$

I hope this clarifies things!

--Chris

9. I'm lost...

I suck at algebra it's due at midpoint

Id pay you if I had money bro

why I go online because the tutors suck and I dont get it at school

10. 36x^2+25y^2=900
5y+6x=30

the solutions are????

11. Originally Posted by Dillon658
I'm lost...

I suck at algebra it's due at midpoint

Id pay you if I had money bro

why I go online because the tutors suck and I dont get it at school
MHF is a community where you get FREE help. We will not take money from you to help you.

I will work them out step by step and hopefully then, it will make sense.

You were asked to find the coordinate $(x,y)$, given the midpoint and one endpoint.

We came up with the formula $\left(\frac{x+19}{2},\frac{y-1}{2}\right)=(27,1)$

We then compare the coordinates. We have an equation for x and y.

$\frac{x+19}{2}=27$

Multiply both sides by 2:

${\color{red}2}\cdot\frac{x+19}{2}={\color{red}2}\c dot 27\implies {\not 2}\cdot\frac{x+19}{\not 2}=54\implies x+19=54$

Subtract 19 from both sides to solve for x:

$x+19{\color{red}-19}=54{\color{red}-19}\implies x+\not 19-\not 19=35\implies \color{red}\boxed{x=35}$

$\frac{y-1}{2}=1$

Multiply both sides by 2:

${\color{red}2}\cdot\frac{y-1}{2}={\color{red}2}\cdot 1\implies {\not 2}\cdot\frac{y-1}{\not 2}=2\implies y-1=2$

Add 1 to both sides to solve for y:

$y-1{\color{red}+1}=1{\color{red}+2}\implies x-\not 1+\not 1=3\implies \color{red}\boxed{y=3}$

Does this make sense?

--Chris

12. Originally Posted by Dillon658
36x^2+25y^2=900
5y+6x=30

the solutions are????

$36x^2+25y^2=900$

$6x+5y=30$

Note that $36x^2+25y^2=900\implies (6x)^2+(5y)^2=900$

To find what x and y are, we need to substitute $6x+5y=30$ into $(6x)^2+(5y)^2=900$...but how? We need to solve for either $6x$ or $5y$.

I'll solve for 5y:

$6x+5y=30\implies 5y=30-6x$

Substitute this into the other equation:

$(6x)^2+(5y)^2=900\implies 36x^2+(\underbrace{30-6x}_{5y})^2=900$

Now expand and simplify.

$36x^2+(900-180x+36x^2)=900\implies 72x^2-180x=0\implies 18x(4x-10)=0$ $\implies x=\tfrac{10}{4} \text{ or }x=0$.

Now find y:

$5y=30-6x\implies 5y=30\text{ or }5y=30-15\implies y=6\text{ or } y=3$

The correct paired solutions would be $\left(\tfrac{10}{4},3\right)$ and $\left(0,6\right)$.

Does this make sense?

--Chris