Determine third point of triangle when two points and all sides are known?
By Sebastian Wright
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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 = 3Can 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.yThanks!
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 = ?8 Answers
$\begingroup$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.
Given coordinates of $A$ and $B$ and lengths $\overline{AB}$, $\overline{AC}$ and $\overline{BC}$ we want to find coordinates of $C$. Let's assume that $A$ is at $(0,0)$ and $B$ is at $(0,\overline{AB})$ i.e. in the $y$-direction. Let's also assume that we know point $C$ is in the positive $x$-direction (there is another solution in the other direction). It should be simple to reduce a more general problem to this case. The solution is then:
$$C_y = \frac {\overline{AB}^2 + \overline{AC}^2 - \overline{BC}^2} {2\overline{AB}}$$
$$C_x = \sqrt{\overline{AC}^2 - {C_y}^2}$$
The derivation of this is given below:
Consider the triangle $ABC$ as two right angle triangles as shown:
Solution of triangle coordinates
If we call $C_x$ and $C_y$ the coordinates of $C$, then Pythagoras theorem for each of the two triangles gives:
$(1) \qquad \overline{AC}^2 = {C_y}^2 + {C_x}^2$
$(2) \qquad \overline{BC}^2 = (\overline{AB}-C_y)^2 + {C_x}^2$
Rearranging $(1)$ gives
$(3) \qquad {C_x}^2 = \overline{AC}^2-{C_y}^2$
Substituting $(3)$ into $(2)$ and rearranging gives
$$\overline{BC}^2 = (\overline{AB}-{C_y})^2 + \overline{AC}^2-{C_y}^2 = \overline{AB}^2 - 2\overline{AB}\cdot{C_y} + \overline{AC}^2$$
$$C_y = \frac{AB^2 + AC^2 - BC^2}{2\overline{AB}}$$
Since $C_y$ is now known $C_x$ can be found by rearranging $(1)$ to give
$$C_x = \pm \sqrt{\overline{AC}^2 - {C_y}^2}$$
$\endgroup$ 2 $\begingroup$Here is a solution in JavaScript
//a,b,c are the sides of the triangle
function get_third_point_coordinates(a, b, c){ var result = {x:0,y:0}; if(a > 0){ result.x = (c*c - b*b + a*a) / (2*a); } result.y = Math.sqrt(c*c - result.x*result.x); return result;
}The triangle is Pythagorean. Since $ab=ch$ the second coordinate of $C$ is $\pm12/5$. From $a^2=cp$ the first coordinate of $C$ is $5/9$ thus there are two solutions.
$\endgroup$ 2 $\begingroup$You can not determine unless there is a direction in the graph.
As i see we have AB,BC,AC. If the way they are written imposes a direction then we see that it is not a directed graph. It would be if the sides would be written AB,BC,CA.
$\endgroup$ 2 $\begingroup$Here's what I've got: X = (Ax+Bx)/2 plus or minus [(1/2)*sqrt(3)*sqrt((Bx-Ax)^2+(By-Ay)^2)]/[sqrt((-(Bx-Ax)/(By-Ay)+1)]
Then solve for y using any random formula such as the distance formula (since you have x)
Where (X,Y) = unknowns, and (Ax, Ay) is one vertex, and (Bx, By) is another.
Note the above equation may be wrong, but how I can tell you how I got it:
I found the midpoint and opposite reciprocal slope of the line of the two existing vertexes. Then, I found the height of the triangle, which is always half of the base * sqrt(3) due to basic trigonometry. Then, using the midpoint, I used the distance formula and the point-slope formula squared to set up an equation for X and Y, which simplified into the answer above.
I'm going to test in a program soon. Just putting my process out there to see if it helps you.
$\endgroup$ 1 $\begingroup$I am attaching a general solution with some R code.
- We need to a system of equations (with two unknowns) that are logically independent.
- The area of a triangle can be computed using the sides (Heron Area) and can also be compared to the area formula using vertices of the triangle.
- The distance formula mentioned above can be used twice (AC and BC) and these equations can be subtracted.
- Understanding the nature of the problem, there will be two solutions: C and C'
heronAreaTriangle = function (AB,BC,AC) { s = 1/2 * (AB+BC+AC); # semi-perimeter Area = sqrt(s * (s-AB) * (s-BC) * (s-AC) ); Area; # returns Area } findTrianglePointC = function(AB,BC,AC, x_a,y_a, x_b,y_b) { M = matrix(0,ncol=2,nrow=2); Area = heronAreaTriangle(AB,BC,AC); dy = y_a-y_b; dx = x_a - x_b; dxy = x_a*y_b - y_a * x_b; y_c = ( dxy + dy * x_c - 2 * Area ) / dx; x_c = ( (BC^2-AC^2) - x_b^2 - y_b^2 + x_a^2 + y_a^2 - 2*dy*y_c) / (2*dx); M[1,1]=x_c; M[1,2]=y_c; y_c = ( dxy + dy * x_c + 2 * Area ) / dx; x_c = ( (BC^2-AC^2) - x_b^2 - y_b^2 + x_a^2 + y_a^2 - 2*dy*y_c) / (2*dx); M[2,1]=x_c; M[2,2]=y_c; result = as.data.frame(M); colnames(result) = c("c_x","c_y"); # result; list("x_a"=x_a,"y_a"=y_a, "x_b"=x_b,"y_b"=y_b, "AB"=AB,"BC"=BC,"AC"=AC, "Area"=Area, "result"=result); }Usage:
findTrianglePointC(AB,BC,AC, x_a,y_a, x_b,y_b);
findTrianglePointC(5,4,3, 0,0, 5,0); # the example above
result = findTrianglePointC(5,4,3, 0,0, 5,0);
result;
result$result;
result$result[1,];
result$result[2,];
as.numeric(result$result[2,]);
result$result[1,]$c_x;
result$result[1,]$c_y;Output:
> findTrianglePointC(5,4,3, 0,0, 5,0); # the example above
$x_a
[1] 0
$y_a
[1] 0
$x_b
[1] 5
$y_b
[1] 0
$AB
[1] 5
$BC
[1] 4
$AC
[1] 3
$Area
[1] 6
$result c_x c_y
1 1.8 2.4
2 1.8 -2.4The logic is correct, although possible errata may exist (I haven't tried it on several cases).
Sources:
- Formula to find the third point of triangle when two points and all sides are known?
- Determine third point of triangle when two points and all sides are known
More verbosity about the logic that is simplified above:
determinateAreaTriangle = function (x_a,y_a, x_b,y_b, x_c,y_c) { # # M = matrix(c(x_a,y_a,1, x_b,y_b,1, x_c,y_c,1), ncol=3); A = abs( 1/2 * det(M) ); A = abs( 1/2 * (x_b*y_c - x_c*y_b - x_a*y_c + x_c*y_a + x_a*y_b - x_b *y_a ) ); # by-hand computation A; # returns Area }
# given points A (x_a,y_a) B (x_b,y_b) and lengths of sides AB, BC, AC # find coordinates for C (x_c, y_c) which should return two possible answers # geometry of two intersection circles from two vertices of triangle = two solutions # heronEquation for area, find numeric value Area = heronAreaTriangle(AB,BC,AC); # use determinateAreaTriangle as an equation to help solve simultaneous system # Area = abs( 1/2 * (x_b*y_c - x_c*y_b - x_a*y_c + x_c*y_a + x_a*y_b - x_b *y_a ) ); # Area = 1/2 * abs ( x_a*y_b - x_b * y_a + x_c * (y_a-y_b) + y_c * (x_b-x_a) ); # 2*Area = abs ( x_a*y_b - y_a * x_b + x_c * (y_a-y_b) - y_c * (x_a - x_b) ); dy = y_a-y_b; dx = x_a - x_b; dxy = x_a*y_b - y_a * x_b; # deltas of x,y, and innerproduct? #2*Area = abs ( dxy + x_c * (dy) - y_c * (dx) ); # y_c = ( dxy + dy * x_c - 2 * Area ) / dx; # y_c = ( dxy + dy * x_c + 2 * Area ) / dx; # distance formula (circle definition) # BC = sqrt( (x_c-x_b)^2 + (y_c-y_b)^2 ); # BC2 = (x_c-x_b)^2 + (y_c-y_b)^2; # squared is circle formula? # AC = sqrt( (x_c-x_a)^2 + (y_c-y_a)^2 ); # AC2 = (x_c-x_a)^2 + (y_c-y_a)^2; # squared is circle formula? # subtract BC2-AC2 # BC2-AC2 = x_b^2 + y_b^2 - x_a^2 - y_a^2 + 2*x_c * (x_a - x_b) + 2*y_c* (y_a-y_b); # BC2-AC2 = x_b^2 + y_b^2 - x_a^2 - y_a^2 + 2*x_c * (dx) + 2*y_c* (dy); # solve for x_c in distance formula (for each y_c) # x_c = ( (BC^2-AC^2) - x_b^2 - y_b^2 + x_a^2 + y_a^2 - 2*dy*y_c) / (2*dx); # # A (-2,2), B (1,5) and C (6,-1) ... maybe C' (-1,13) # x_a = -2; y_a = 2; x_b = 1; y_b = 5; x_c = 6; y_c = -1; # AB=3*sqrt(2); BC = sqrt(61); AC=sqrt(73); # # A (0,0), B (5,0) and C (9/5,12/5) ... maybe C' (9/5,-12/5) # x_a = 0; y_a = 0; x_b = 5; y_b = 0; x_c = 9/5; y_c = 12/5; # AB=5; BC = 4; AC=3; I would keep simple things simple.
Recognize the Pythagorean triplet of sides $(3,4,5)$
Vector approach. Coordinates of C : $(x,y)$.
$$ \vec{AC}\cdot \vec{BC} =0 \tag1$$$$ (xi+yj)\cdot[ (x-5)i+y j=0] \rightarrow \,x^2+y^2-5 x=0 \tag2$$
This is only a locus, and by Thales theorem or should contain both your points at a right angle vertex.
To isolate points of intersection draw a circle of radius $3$ to cut the locus.$$ x^2+y^2= 3^2 \tag3$$
Subtracting equations 2) and 3) we get solutions of quadratic equation which has two real roots.
$$ x= \frac{9}{5}=1.8 \tag4$$
Plug into either of the two equations for example into the second one 3)$$ y^2= 9- (9/5)^2 = \frac{9\times 16}{25}\rightarrow y= \pm2.4$$
Thus two solution points are $$ (1.8,+2.4),\,(1.8,-2.4) \tag5$$
which we can label
$$ C_1, C_2$$
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