Determine third point of triangle (on a 2D plane) when two points and all sides are known?
A = (0,0)
B = (5,0)
C = (?, ?)
AB = 5
BC = 4
AC = 3
Can someone please explain how to go about this? I understand there will be two possible points and would like to arrive at both.
This is what I've worked out but I'm uncertain at this point how correct it is.
C.x = (AB² - BC² + AC²) / (2 * AB)
C.y = sqrt(BC² - (B.x - C.x)²) - B.y
Thanks!
Update – Need to turn the answer into a reusable formula, solving for C.x and C.y
known sides AB, BC, AC
known points A(x, y), B(x, y)
unknown points C(x, y)
AC² - BC² = ((Ax - Cx)² + (Ay - Cy)²) - ((Bx - Cx)² + (By - Cy)²)
Goal:
C.x = ?
C.y = ?
Best Answer
The distance between two points $P(p_1,p_2)$ and $Q(q_1,q_2)$ in the plane is$\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}$.
Let us denote coordinates of $C$ by $(x,y)$. Then the distances between from $C$ to $A$ and from $C$ to $B$ will be $$\sqrt{x^2+y^2}=3 \Rightarrow x^2+y^2=9$$ $$\sqrt{(x-5)^2+y^2}=4\Rightarrow (x-5)^2+y^2=16$$ respectively. Substracting the first equation from the second equation we get $$(x-5)^2-x^2=7 \Rightarrow x^2-10x+25-x^2=7$$ $$\Rightarrow -10x+25=7 \Rightarrow x=\frac{-18}{-10}=\frac{9}{5}$$ Now if we substitute $x=\frac95$ into one of the above equations we get $y=\pm \frac{12}5$ So we find $(x,y)=(\frac95,\frac{12}{5})$ and $(x',y')=(\frac95,-\frac{12}{5})$.
As you see from the picture these two points are symmetric w.r.t $-x$ axis.