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.
help pls
Are these like x,y,z coordinates?
yes
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.\[d = \sqrt {\Delta^2 x + \Delta^2 y + \Delta^2 z} = \sqrt {(x_2-x_1)^2 + ... }\]\[A = \frac {1}{4} \sqrt { (a^2 + b^2 + c^2)^2 - 2(a^4 + b^4 + c^4)}\]
but rogue there are only 2 points read carefully
Oh, so its just x,y coordinates?
If so, then you can still use the same method I said, only theres no delta^z in the distance formula.
phi wats ur opinion?
dumbcow can u hlp?
so this is in xy plane? yeah use distance formula to get side lengths
Then to get height, take one side and find line perpendicular to it that goes through opposite vertex
wat abt taking determinants?
yeah there prob is a way to get area using matrices...i don't remember how though
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)
herons formula should work ok
Area = base*height find where lines intersect, use those points to get side lengths
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
how can a line intetsect all 3 sides of a triangle ??
DONT KNOW
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)
Here is how I would do it
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