## nagdy1 ABC has vertices A(0, 6), B(4, 6), and C(1, 3). Sketch a graph of ABC and use it to find the orthocenter of ABC. Then list the steps you took to find the orthocenter, including any necessary points or slopes you had to derive. one year ago one year ago

1. jennychan12

too many, ugh geometry. sorry not my strength,

2. nagdy1

huh

3. jackxie14

will do u no how to graph??

4. nagdy1

i will sketch wait

5. Johnjakile1998

Do you know what you're looking for?

6. tcarroll010

The orthocenter is where the altitudes all intersect, altitudes being from a vertex tdrawn perpendicular to the opposite side. You only have to get 2 altitudes and that intersection will automatically be where the thirs side altitude will interesect. So, you can get the equation of a line through a vertex perpendicular to the opposite side. Do that again for another point. To get perpendicularity, use the slope formula for 2 points. You will get 2 lines that intersect at a point, the orthocenter. This outline how to do it algebraically.

7. jackxie14

ok:)

8. forty-two

I just have to say, haha, that was awesome, now you have like a billion people looking at your question haha

9. nagdy1

10. tcarroll010

So, for the slope of 2 points, A and B, (0, 6) and (4, 6), use:$m = \frac{ y _{1} - y _{2} }{ x _{1} - x _{2} }$where "m" is your slope and (x1, y1) and (x2, y2) are the 2 points A and B. Take the negative reciprocal of this slope for one of the 2 lines you want to get and use the point-slope formula with "C" as your new (x1, y1):$y - y _{1} = m(x - x _{1})$And then convert this to the slope-intercept form: y = mx + b Repeat the process for another vertex, say "A" (or "B", doesn't matter). You now have 2 lines that intersect (independent system) and that intersection will be the orthocenter.

11. nagdy1

can you show me the equation then ill solve it

12. jackxie14

true @forty-two :)

13. tcarroll010

np. The first formula is in my last post (first formula) and just use points A and B. Get the slope and then the negative reciprocal. The negative reciprocal will be -1/m, once you get "m".

14. nagdy1

15. tcarroll010

So, x1=0, y1=6, x2=4, and y2=6

16. nagdy1

ok wait

17. tcarroll010

m = (6 - 6)/(0 - 4) = 0 and we have a horizontal line for those two points A and B. In this case, instead of the negative reciprocal, we have an undefined slope or a vertical line from "C" for the altitude.

18. nagdy1

so the answer is undefined slope or a vertical line from "C" for the altitude.

19. tcarroll010

Now, one could say, "Hey wait, this is going to be complicated because our slopes are 0 and "undefined"". That's ok. We can still use the point-slope formula, knowing that we will be using a vertical line for the altitude.

20. tcarroll010

It actually becomes easier in this case, because our first line, our first altitude, is going to be a vertical line through "C". This is the equation x= 1. So we actually "luck out" from what was at first going to be looking like a complication. It becomes a simplification.

21. tcarroll010

So, 1 altitude is done and with a very simple equation for its line. We can already anticipate solving the linear system (with the other altitude) by a very simple substitution.

22. tcarroll010

So, do you want to take a crack at getting the other altitude? It would be tremendous practice for anyone and it would consolidate your understanding. Want a go?

23. nagdy1

can u show me the answer

24. tcarroll010

(Sigh) I'll show you the work: Take points B and C for your new side and A as your vertex for the altitude. m = (6 - 3)/(4 - 1) = 1. We want the negative reciprocal which is -1 Point-slope: y - 6 = (-1)(x - 0) -> y = -x + 6 is the equation for the second altitude.

25. tcarroll010

We resolve this with x = 1 -> y = -(1) + 6 -> y = 5 Orthocenter: (1, 5)

26. nagdy1

thanx

27. tcarroll010

It would be a good idea to get this same answer by a second way (1, 5) by using other pairs of points, where you are not dealing with slopes 0 and "undefined". Then, if you understand how to do that yet another way, then you will always be able to do this type of problem on your own. At the very least, re-read everything I wrote down here and you'll probably still understand it, but the best way is by practice.

28. tcarroll010

You're welcome.