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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
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

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jennychan12Best ResponseYou've already chosen the best response.1
too many, ugh geometry. sorry not my strength,
 one year ago

jackxie14Best ResponseYou've already chosen the best response.0
will do u no how to graph??
 one year ago

Johnjakile1998Best ResponseYou've already chosen the best response.0
Do you know what you're looking for?
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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.
 one year ago

fortytwoBest ResponseYou've already chosen the best response.0
I just have to say, haha, that was awesome, now you have like a billion people looking at your question haha
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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 pointslope formula with "C" as your new (x1, y1):\[y  y _{1} = m(x  x _{1})\]And then convert this to the slopeintercept 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.
 one year ago

nagdy1Best ResponseYou've already chosen the best response.0
can you show me the equation then ill solve it
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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".
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
So, x1=0, y1=6, x2=4, and y2=6
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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.
 one year ago

nagdy1Best ResponseYou've already chosen the best response.0
so the answer is undefined slope or a vertical line from "C" for the altitude.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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 pointslope formula, knowing that we will be using a vertical line for the altitude.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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?
 one year ago

nagdy1Best ResponseYou've already chosen the best response.0
can u show me the answer
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
(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 Pointslope: y  6 = (1)(x  0) > y = x + 6 is the equation for the second altitude.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
We resolve this with x = 1 > y = (1) + 6 > y = 5 Orthocenter: (1, 5)
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.4
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, reread everything I wrote down here and you'll probably still understand it, but the best way is by practice.
 one year ago
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