# How to get the third point coordinates in isosceles triangle?

Isosceles triangle $ABC$
$AB = AC = d_1$
$BC = d_2$

$A = (x_1, y_1)$ $B = (x_2, y_2)$ $C = (x_3, y_3)$

$\angle BAC = \phi$
$\angle ABC =\angle ACB = \theta$

I want an equation for $x_3$ and $y_3$ (and I know there will be two values)

• Hint $$y_2=y_3$$ $$x_1=\frac{x_2+x_3}{2}$$ – Shabbeh Jul 28 '14 at 2:59
• @Vincent I searched and found similar questions but couldn't find the exact one nor the wanted answer, I've been trying for about 4 hours but couldn't find it yet.. I tried using Pythagorean theorem, law of sines and law of cosines... – Aly Elhaddad Jul 28 '14 at 3:04
• I don’t quite understand what you are asking. The following are my guesses. (1) $d_1$, $d_2$, ϕ, and θ are the givens; (2) want to find $x_3$ (also $y_3$) in terms of the givens. If they are true, I have the following comments:-  either ϕ or θ is known will be sufficient because one is derivable from the other.  d1 and d2 are also inter-derivable with the help of θ (or ϕ).  Do a series of transformations, to reduce your triangle to the one being symmetrically located about the y-axis and the midpoint of BC at the origin. Then $y_3 = 0$, and $x_3 = - x_2 = d_1 cos θ$. – Mick Jul 28 '14 at 5:39
• @Mick , You guessed right. About your third comment, can you clarify? – Aly Elhaddad Jul 28 '14 at 17:57
• @AlyEl-Haddad Transformations include translate [or commonly called shift] (left/right/up/down); rotate (through certain angles); reflect (about an axis). These transformations are shape and size preserving. Any object, after going through a combination of these transformations, can be “reduced” to a more convenient location. In this way, the original problem can be reduced to a much more simplified equivalent (like those I have suggested). If the reduced version can be solved, the original problem is then considered as solved also. – Mick Jul 29 '14 at 1:51

## 3 Answers

I could solve this. The solution key was the conversion between the Cartesian Coordinate System and the Polar Coordinate System.

$$x_3 = AB * Cos(ϕ) + x_1$$

$$y_3 = AB * Sin(ϕ) + y_1$$

If you have all the sides, you don't need the angles-which do you believe? Taking the sides, you have two equations in two unknowns: one for the distance from $A$ and one from the distance from $B$. So $(x_3-x_1)^2+(y_3-y_1)^2=d_1^2$ and $(x_3-x_2)^2+(y_3-y_2)^2=d_2^2$ You are correct there will be two solutions. With two quadratics you might expect four, but two will be extraneous.

• thanks but I could find this already find this I need something like x_3$= ... y_3$ = ... – Aly Elhaddad Jul 28 '14 at 3:19

Since AC = $$d_1$$,
$$d_1^{2}$$ = $$(x_3-x_1)^2$$ + $$(y_3-y_1)^2$$ $$\ \ \ \ \ \ \ \$$ _$$1$$

And since BC = $$d_2$$
$$d_2^{2}$$ = $$(x_3-x_2)^2$$ + $$(y_3-y_2)^2$$ $$\ \ \ \ \ \ \ \$$ _$$2$$

Now there are two unknowns and two equations.
Using these two equations, we can get $$x3$$ and $$y3$$.

$$y_3 = y_1 \pm\sqrt{d_1^2-(x_3-x_1)^2}$$

Substituting this value of $$y_3$$ from $$eqn_1$$ into $$eqn_2$$, we can get $$x_3$$.

$$x_3 = \dfrac{-4 d_1^2 x_1 + 4 d_1^2 x_2 \pm \sqrt{(4 d_1^2 x_1 - 4 d_1^2 x_2 - 4 d_2^2 x_1 + 4 d_2^2 x_2 - 4 x_1^3 + 4 x_1^2 x_2 + 4 x_1 x_2^2 - 4 x_1 y_1^2 + 8 x_1 y_1 y_2 - 4 x_1 y_2^2 - 4 x_2^3 - 4 x_2 y_1^2 + 8 x_2 y_1 y_2 - 4 x_2 y_2^2)^2 - 4 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2) (d_1^4 - 2 d_1^2 d_2^2 - 2 d_1^2 x_1^2 + 2 d_1^2 x_2^2 - 2 d_1^2 y_1^2 + 4 d_1^2 y_1 y_2 - 2 d_1^2 y_2^2 + d_2^4 + 2 d_2^2 x_1^2 - 2 d_2^2 x_2^2 - 2 d_2^2 y_1^2 + 4 d_2^2 y_1 y_2 - 2 d_2^2 y_2^2 + x_1^4 - 2 x_1^2 x_2^2 + 2 x_1^2 y_1^2 - 4 x_1^2 y_1 y_2 + 2 x_1^2 y_2^2 + x_2^4 + 2 x_2^2 y_1^2 - 4 x_2^2 y_1 y_2 + 2 x_2^2 y_2^2 + y_1^4 - 4 y_1^3 y_2 + 6 y_1^2 y_2^2 - 4 y_1 y_2^3 + y_2^4)} + 4 d_2^2 x_1 - 4 d_2^2 x_2 + 4 x_1^3 - 4 x_1^2 x_2 - 4 x_1 x_2^2 + 4 x_1 y_1^2 - 8 x_1 y_1 y_2 + 4 x_1 y_2^2 + 4 x_2^3 + 4 x_2 y_1^2 - 8 x_2 y_1 y_2 + 4 x_2 y_2^2}{2 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2)}$$

Similarly,
$$x_3 = x_1 \pm\sqrt{d_1^2-(y_3-y_1)^2}$$

Substituting this value of $$x_3$$ from $$eqn_1$$ into $$eqn_2$$, we can get $$y_3$$.

$$y_3 = \dfrac{-4 d_1^2 y_1 + 4 d_1^2 y_2 \pm \sqrt{(4 d_1^2 y_1 - 4 d_1^2 y_2 - 4 d_2^2 y_1 + 4 d_2^2 y_2 - 4 x_1^2 y_1 - 4 x_1^2 y_2 + 8 x_1 x_2 y_1 + 8 x_1 x_2 y_2 - 4 x_2^2 y_1 - 4 x_2^2 y_2 - 4 y_1^3 + 4 y_1^2 y_2 + 4 y_1 y_2^2 - 4 y_2^3)^2 - 4 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2) (d_1^4 - 2 d_1^2 d_2^2 - 2 d_1^2 x_1^2 + 4 d_1^2 x_1 x_2 - 2 d_1^2 x_2^2 - 2 d_1^2 y_1^2 + 2 d_1^2 y_2^2 + d_2^4 - 2 d_2^2 x_1^2 + 4 d_2^2 x_1 x_2 - 2 d_2^2 x_2^2 + 2 d_2^2 y_1^2 - 2 d_2^2 y_2^2 + x_1^4 - 4 x_1^3 x_2 + 6 x_1^2 x_2^2 + 2 x_1^2 y_1^2 + 2 x_1^2 y_2^2 - 4 x_1 x_2^3 - 4 x_1 x_2 y_1^2 - 4 x_1 x_2 y_2^2 + x_2^4 + 2 x_2^2 y_1^2 + 2 x_2^2 y_2^2 + y_1^4 - 2 y_1^2 y_2^2 + y_2^4)} + 4 d_2^2 y_1 - 4 d_2^2 y_2 + 4 x_1^2 y_1 + 4 x_1^2 y_2 - 8 x_1 x_2 y_1 - 8 x_1 x_2 y_2 + 4 x_2^2 y_1 + 4 x_2^2 y_2 + 4 y_1^3 - 4 y_1^2 y_2 - 4 y_1 y_2^2 + 4 y_2^3}{2 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2)}$$

Links to solutions

$$x3$$:
http://www.wolframalpha.com/input/?i=solve+d2%5E2%3D(x3-x2)%5E2+%2B+(sqrt(d1%5E2+-+(x3+-+x1)%5E2)+%2B+y1+-+y2)%5E2+for+x3

$$y3$$:
http://www.wolframalpha.com/input/?i=solve+d2%5E2%3D(sqrt(d1%5E2+-+(y3+-+y1)%5E2)+%2B+x1+-x2)%5E2+%2B+(y3+-+y2)%5E2+for+y3