0
$\begingroup$

I want to plot a triangle, given side lengths $a$, $b$, and $c$.

I can plot point $A$—opposite side $a$—at the origin $(A_x = 0,\ A_y = 0)$.

I can plot point $B$—opposite side $b$—along the $x$-axis at $(B_x = c,\ B_y = 0)$.

My goal is to solve for the coordinates $(C_x,\ C_y)$ for any triangle. I understand there will be two possible solutions: one where the triangle “points up” and one where the triangle “points down”. I am only interested in the solution where $C_y$ is positive.

Please provide a solution in the form:

$C_x =$

$C_y =$

$\endgroup$
4
  • $\begingroup$ With a ruler and a compass :o) $\endgroup$ Commented Dec 20, 2015 at 15:29
  • $\begingroup$ You have a bit of a mistake: point B, if on the $X$ axis, should be at $(b,0)$ not $(c,0)$. $\endgroup$ Commented Dec 20, 2015 at 15:33
  • $\begingroup$ @MarkFischler I think that the side $c$ is along the $x$-axis, so this point would be correct. $\endgroup$ Commented Dec 20, 2015 at 15:34
  • $\begingroup$ @michael Burr - you are right. Hint: place point $C$ at $(x,y)$. Then the condition that side $a$ is of length $a$ gives $x^2 - 2cx +c^2 + y^2 = a^2$. The condition that side $b$ is of length $b$ gives a simpler equation: $x^2 + y^2 = b^2$. Subtract to get a linear equation for $x$. $\endgroup$ Commented Dec 20, 2015 at 15:42

3 Answers 3

Reset to default
3
$\begingroup$

The points $C_x$ and $C_y$ satisfy the following system of equations:

\begin{align} \sqrt{C_x^2+C_y^2}&=b\\ \sqrt{(c-C_x)^2+C_y^2}&=a \end{align}

Square both equations and substitute so that you are left with a linear expression for $C_x$. In other words,

\begin{align} C_x^2+C_y^2&=b^2\\ c^2-2cC_x+C_x^2+C_y^2&=a^2 \end{align}

Then, by substitution and rearranging, you have that $$ C_x=\frac{a^2-b^2-c^2}{-2c} $$

Then, substitute and solve for $C_y$. $$ C_y=\sqrt{b^2-C_x^2}. $$

$\endgroup$
0
1
$\begingroup$

By the law of cosines, where $\alpha$ is the triangle at point $A$,

$$a^2=b^2+c^2-2bc\cos\alpha$$ so $$\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$$

Then, since side $b$ is the hypotenuse of a right triangle with leg $C_x$ adjacent to angle $\alpha$,

$$C_x=b\cos\alpha=\frac{b^2+c^2-a^2}{2c}$$

Since the other leg of that right triangle is $C_y$, by the Pythagorean theorem we get

$$C_y=\sqrt{b^2-C_x^2}=\sqrt{b^2-\left(\frac{b^2+c^2-a^2}{2c}\right)^2}$$

If you think that formula for $C_y$ is too messy, you could find another one by using two formulas for the area of the triangle, the second given by Heron's formula. Here $s=\dfrac{a+b+c}{2}$ is the semiperimeter of the triangle.

$$\frac 12bC_y=\sqrt{s(s-a)(s-b)(s-c)}$$ so

$$C_y=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{b}$$

As a bonus, the formula for $C_x$ gives you the cosine of the angle at $A$ and the second formula for $C_y$ gives you the area of the triangle.

$\endgroup$
1
  • 2
    $\begingroup$ I'm curious, why the downvote? All the information in this post is correct, I derive it by means that are different from and shorter than the accepted answer, I give alternative formulas, and I note that other useful information can be easily derived along the way. I understand it if you prefer another answer, but why not just upvote that other answer? $\endgroup$ Commented Jan 3, 2016 at 13:04
0
$\begingroup$

if you know point $a,b$ then lets assume $C(x,y)$ so by distance formula you'll get two equations solve for indivdual $x,y$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.