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Mathematics 13 Online
OpenStudy (aravindg):

show that area of the triangle with the vertices [(a+1)(a+2),a+2],[(a+2),(a+3),a+3] and [(a+3)(a+4),a+4]is independent of a.

OpenStudy (aravindg):

help pls

OpenStudy (rogue):

Are these like x,y,z coordinates?

OpenStudy (aravindg):

yes

OpenStudy (rogue):

Hmm, I'm not really sure how to do this, but heres an idea. Use the 3D distance formula to find the length between the 3 points, than use heron's formula to find the area. It might be really messy, so there is probably an easier way.\[d = \sqrt {\Delta^2 x + \Delta^2 y + \Delta^2 z} = \sqrt {(x_2-x_1)^2 + ... }\]\[A = \frac {1}{4} \sqrt { (a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}\]

OpenStudy (aravindg):

but rogue there are only 2 points read carefully

OpenStudy (rogue):

Oh, so its just x,y coordinates?

OpenStudy (rogue):

If so, then you can still use the same method I said, only theres no delta^z in the distance formula.

OpenStudy (aravindg):

phi wats ur opinion?

OpenStudy (aravindg):

dumbcow can u hlp?

OpenStudy (dumbcow):

so this is in xy plane? yeah use distance formula to get side lengths

OpenStudy (dumbcow):

Then to get height, take one side and find line perpendicular to it that goes through opposite vertex

OpenStudy (aravindg):

wat abt taking determinants?

OpenStudy (dumbcow):

yeah there prob is a way to get area using matrices...i don't remember how though

OpenStudy (aravindg):

show that area of parallelogram formed by lines 3y-2x=a ,2y-3x+a=0,2x-3y+3a=0 and 3x-2y-2a=0 is (2a^2/5)

OpenStudy (dumbcow):

herons formula should work ok

OpenStudy (dumbcow):

Area = base*height find where lines intersect, use those points to get side lengths

OpenStudy (aravindg):

a line L intersects the three sides BC,CA and AB of a triangle ABC at P,Q and R .then show that (BP/BC).(CQ/QA).(AR/RB)=-1

OpenStudy (dumbcow):

how can a line intetsect all 3 sides of a triangle ??

OpenStudy (aravindg):

DONT KNOW

OpenStudy (phi):

You can use the area of a triangle defined by 3 points (x1,y1), (x2,y2), (x3,y3) is given by \[A=\frac{1}{2}\left| \left[\begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix}\right] \right|\] btw, the vertices [(a+1)(a+2),a+2],[(a+2),(a+3),a+3] and [(a+3)(a+4),a+4] has a typo, in that the second point is [[(a+2)(a+3),a+3] (only 2 points, obviously)

OpenStudy (phi):

Here is how I would do it

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