hard geometry/cartesian plane question!?
A triangle is made up of three points on a cartesian plane. One of the sides has endpoints (-1,-1) and (3,5). To the right of this side are the two other sides. The one on top is 6 units long and the bottom one is 3 units long. What is the coordinate of the missing point? I know how to find the length of the missing side and all the angles but I don't know how to find the point. Thanks for you help!
Update:Ah wow wasn't thinking in terms of circles. Thanks for the help!
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let, (x,y) is the required point.
(x+1)^2+(y+1)^2=9............(1)
(x-3)^2+(y-5)^2=36............(2)
subtracting (2) from (1),
8x+12y=5
or, y=(5-8x)/12.......(3)
putting (3) in (1),
(x+1)^2+((5-8x)/12+1)^2=9
after required steps,
208x^2+16x-863=0
so, x=(-16+(16^2+4*863*208)^(1/2))/(2*208) or (-16-(16^2+4*863*208)^(1/2))/(2*208)
from graph, its absurd for x to be negative.
so, x=(-16+(16^2+4*863*208)^(1/2))/(2*208)
=1.9988204
from (3),
y=-.91588
so, (x,y)=(1.9988204,.91588)
As per your statement the distance 6, squared 36 = (x--3)^2 + (y--5)^2-->(1)
Same way the other length, 3, squared 9 = (x+1)^2 + (y+1)^2--->(2)
Here (x,y) is the unknown co ordinates of the third point.
Solving (1) and (2)
(2) multiplied by 4 makes it equal to (1)
So 4(x+1)^2 + 4(y+1)^2 = (x--3)^2 + (y--5)^2
3x^2 +3y^2 +14x +18y --26 = 0
This will be an equation of a circle, the point on which would fulfill your condition that the respective lengths are 6 and 3 units. Hence so many points!