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I am really having problems solving the following problems:

  1. If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the two vertices on the hypotenuse of a right angled isosceles triangle then the coordinates of the 3rd vertex are?
  2. Similarly-If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the two vertices of an Equilateral Triangle then the coordinates of the 3rd vertex are?

The answer to 1 and 2 respectively are-

$$\left(\frac{x_1+x_2 \pm(y_1-y_2)}{2},\frac{y_1+y_2 \pm(x_1-x_2)}{2}\right)$$

And

$$\left(\frac{x_1+x_2 \pm\sqrt 3(y_1-y_2)}{2},\frac{y_1+y_2 \pm\sqrt 3(x_1-x_2)}{2}\right)$$

How do I solve this problem?

I tried attempting it by using the fact that the triangle is isosceles and right angled in the first case. So I applied pythagoras theorem but failed to get this answer it was a complicated equation. Secondly I tried multiplying the slopes of the legs of the triangle to get -1 but even this attempt failed.

It was similar with the equilateral triangle problem.

I guess if I understand how to do even one easily the second should bit be difficult.

Please help.

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  • $\begingroup$ Do the consecutive + and - stand for plus-minus? And are you familiar with complex numbers? $\endgroup$ – G-man Apr 1 '15 at 20:09
  • $\begingroup$ Yes they do.. Plus or minus.. Yes i am familiar $\endgroup$ – Jai Mahajan Apr 1 '15 at 20:13
  • $\begingroup$ I suggested an edit. Did I screw up anything? $\endgroup$ – G-man Apr 1 '15 at 20:25
  • $\begingroup$ Just the brackets now it fine.. The whole thing is divided by 2 $\endgroup$ – Jai Mahajan Apr 1 '15 at 20:26
  • $\begingroup$ CAN this question be done Without Rotation or complex nos using simple pythogoras theorem or distance formula maybe? $\endgroup$ – Jai Mahajan Apr 2 '15 at 8:59
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Consider the fact that if P(x,y) is the unknown point, then when you apply rotation around that point, the point P1(x1, y1) goes to (gets mapped to) P2(x2,y2). From this you can construct equations and solve them to get the x and y. Just lookup the formulas for rotation in the 2D plane. The rotation angles are nice in both cases: 90 and 60 degrees. If you do this, you should get the answers you have posted here.

Say you rotate the point (x1,y1) around the point (a,b) on an angle of $\theta$.
Then the point (x1,y1) gets mapped to (x2, y2) where:
$x_2 = (x_1-a) * cos(\theta) - (y_1-b) * sin(\theta) + a$
$y_2 = (x_1-a) * sin(\theta) + (y_1-b) * cos(\theta) + b$

In your case you don't know the $a$ and $b$, you want to find them. The $\theta$ is $\pm \pi/2$ and $\pm \pi/3$ respectively. The $x_1, y_1, x_2, y_2$ - these you know.

Here is how one can solve the case: $\pi/3$. You have 3 more cases to solve. enter image description here

See also:

Rotation
Rotation matrix

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  • $\begingroup$ The answers you have given isnt correct.. Please read my question and the answer again $\endgroup$ – Jai Mahajan Apr 1 '15 at 20:12
  • $\begingroup$ No, thanks. You read, think, and calculate some more. Using my formulas I got your answers using paper and pencil. $\endgroup$ – peter.petrov Apr 1 '15 at 20:20
  • $\begingroup$ Post the pic of your working then please. $\endgroup$ – Jai Mahajan Apr 1 '15 at 20:22
  • $\begingroup$ Posted, you should say thanks a lot. $\endgroup$ – peter.petrov Apr 1 '15 at 20:41
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    $\begingroup$ It will work too for the right angled triangle, just try it yourself. What about the plus/minus: well, they correspond to your unknown point being "above" or "below" the segment formed by the 2 known points. That's why I said you need to solve 3 more cases for theta: (1) -pi/3, (2) pi/2, (3) -pi/2. $\endgroup$ – peter.petrov Apr 1 '15 at 21:25

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