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Um, can someone please help me with proving the collinear lines here? I'm stuck. Thanks

Good job AELN   !!!!!

And.....welcome aboard   !!!!!!

3x+15 = y    sub that in to the circle equation

x^2   +  [(3x+15)/4]^2 = 36

x^2 + [(9x^2 +90x+225)/16 ] = 36

16x^2  + 9x^2 + 90x + 225 = 576

25x^2 + 90x - 351 = 0       Use quadratic formula to find     x  =  2.35692    and  −5.95692      (these are  9/5 +- 12 sqrt5/5 )

Now you can use these values of x to find the y values by substituting into the line equation

   THEN you can use the distance formula to calculate the length    sqrt  { (x1-x2)^2  + (y1-y2)^2 }

Is there an easier way???  Probably........

p + q + r  =   -3

px + qy  + rz    =  1

p + qx + ry  = z

pz + q + rx  = y

py + qz + r  =  x

Add the last  three equations  and we  get that

(p + q + r)  +  ( q + r)x  +  (p + r)y  + (p + q)z  = x + y + z

 -3  +  (q + r)x  + (p + r)y   + (p + q)z   = x + y + z          

Manipulating the first equation  and  substituting we  have

-3  +  (-3-p)x  + (-3 -q)y  + (-3  -r)z  = x + y + z

-3    + -3x  - px  + -3y  -qy  + -3z - rz   =  x + y + z          rearrange  as

-3  - px   - qy   -  rz   =     x + 3x  + y + 3y  + z + 3z

-3  - ( px + qy  + rz) = 4x + 4y + 4z

 -3  -   (1)  =    4 ( x + y + z)

-2  =  4 ( x + y + z)          divide  both sides by 4

-2/4   = x + y + z

-1/2  = x + y + z

I still want to think about it some more......

Madyl wants this one answered.

Ok, got that! Thank you, I am new to this site. However, you should just go around being disrespectful to people

Thumbs up = add a point

Thumbs down = subtract a point.    It should rarely be used.

Flagging sends a message to moderators and should not be used just because you do not like, or do not get, an answer.

Flagging can be used if you believe a post should be removed.

What, does flagging take points off? In Brainly, they recommend flagging!

ok, you do that.

If you didn't understand why it was not complete you should have asked for clarification.

It is still not complete. Don't you have the original question?

If you take points from people who offer assistance, even if the help is just pointing out that the question is incomplete, then you do need someone to tell you to behave.

Thank you! :)

Thank you so much!!! ^^

Thank! ►

Awesome, dude!

Well do your clothes count as bags? Hmmmmmm... or maybe he is referring to our epidermis as the baggy material?

I'm not sure which problem to solve! There are two, and the links lead to links that lead to other links! Can you proivde the original question? I'm confused now.

We can use the Law of Cosines

BC   =  √ [ 13^2  + 15^2  - 2(13 * 15)  cos (95°) ]   ≈   20.7

Cheats? How do you cheat on this?

Coolio!

Yeah, we are FLESH! We don't have bags around us!

Let triangle EDF  be  medial, then  AE  = EC   and  that  BD = DC

So....ED splits the sides  AC and BC  in the same proportion  1 : 1

So .....it is parallel to AB.....

And  since  AE  =(1/2) AC  and  BD  = (1/2) BC.....then  ED  =(1/2) AB

And  by the  same reasoning, we  can show that  EF  =(1/2)BC and FD = (1/2) AC.....then triangles  ABC and DEF are similar figures......and similar figures are  similar in all respects.....so altitude of  medial triangle DEF  = (1/2) altitude of triangle ABC