Thread: Determining the third point on an isosceles triangle

1. Determining the third point on an isosceles triangle

Given the x/y values of the two points on the base, and the length of the altitude -- or 'height' -- of the triangle, is it possible to determine the x,y value of the third point? Thanks in advance for any ideas.

2. Yes!

3. Ok, I suppose I need to elaborate...what would the process be to determine the x/y value of the third point? I read somewhere about finding the midpoint and then using a unit vector, but I'm not clear on how that would look (if that's a viable solution)

4. Hello, MrGreenjeens!

Given the coordinates of the two points on the base,
and the length of the altitude -- or 'height' -- of the triangle,
is it possible to determine the coordinates of the third point?

Yes, and there will be two possible locations for the third vertex,
one above the base and one below the base.

Code:
              R
(x,y)
*
/|\
/ | \
/  |  \
r/   |h  \r
/    |    \
/     |     \
/      |      \
P * - - - * - - - * Q
(x1,y1)    M    (x2,y2)

$\text{The base vertices are: }\:P(x_1,y_1)\text{ and }Q(x_2,y_2)$

$\text{Their midpoint is: }\:M\left(x_m,\,y_m) \:=\:\left(\tfrac{x_1+x_2}{2},\:\tfrac{y_1+y_2}{2}\right)$

$\text{The third vertex is: }\,R(x,y) \,\text{ and }|M\!R| = h$

$\text{Let }\vec u \:=\:\overrightarrow{PQ} \:=\:\langle x_2\!-\!x_1,\:y_2\!-\!y_1\rangle$

$\text{Let }\vec v \:=\:\overrightarrow{MR}\,\text{ where }\vec v \perp \vec u$

. . . $\text{Then: }\:\vec v \:=\:\langle y_1\!-\!y_2,\:x_2\!-\!x_1\rangle$

$\text{The unit vector in the direction of }\vec v\text{ is:}$

. . $\vec w \;=\;\left\langle \frac{y_1-y_2}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}},\; \frac{x_2-x_1}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}}\right \rangle$

$\text{Hence: }\:\vec R \;=\;\vec M + h\vec w$

$\text{Therefore: }\;R\left(\frac{x_1\!+\!x_2}{2} + \frac{h(y_1\!-\!y_2)}{\sqrt{(x_2\!-\!x_1)^2+(y_2\!-\!y_1)^2}}, \;\; \frac{y_1\!+\!y_2}{2} + \frac{h(x_2\!-\!x_1)}{\sqrt{(x_2\!-\!x_1)^2 + (y_2\!-\!y_1)^2}} \right)$

5. Thanks, Soroban -- this is exactly what I was looking for. I had thought that there would be two possible points, but from playing around with your formula, it seems that this is determined by whether the value of h is positive or negative. Anyways, I've been able to integrate this solution into a program I'm working on and it's working as desired, so I appreciate the time you spent.

,

,

,

how to find the third vertex of an isosceles triangle

Click on a term to search for related topics.