2010-11-26, 02:24 PM
Yeah, you're given that the base (DQ) is 2k/3. So you need to figure out the height, using a x/y axis system, I --- just found out by writing this that I screwed up xD, I think I know how to do it now.
Wheee got it
I used a system on x/y axis.
basically:
AQ segment is y=-3/2x + k
and DP is y=3/4x
you put it as y=y and get M (x,y)
3/4x = -3/2x + k
therefor x = 4k/9, replace in either equation and y=k/3
Y in M is the height of the triangle
base * height /2
2k/3 * k/3 = 2k^2/9 divide by 2 and you get the result k^2/9
Wheee got it
I used a system on x/y axis.
basically:
AQ segment is y=-3/2x + k
and DP is y=3/4x
you put it as y=y and get M (x,y)
3/4x = -3/2x + k
therefor x = 4k/9, replace in either equation and y=k/3
Y in M is the height of the triangle
base * height /2
2k/3 * k/3 = 2k^2/9 divide by 2 and you get the result k^2/9