How to find co ordinates of a triangle after increasing the area by a factor of $\alpha$?

Question

i am given with a triangle $\{(x_1,y_1),(x_2,y_2),(x_3,y_3)\}$ and the area need to be increased by a factor $\alpha$. can i anyone let me know formula to find the co ordinates of new triangle? There are many possibilities but i want something as follows.

Answer 1

Well, in general, you can multiply all the coordinates by $\sqrt{a}$ and you have a triangle with $a$ times the area. But from the figure, I guess you need something like this.

First, move the center of mass of your triangle to the origin, i.e. calculate $(x_0, y_0) = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$, and then consider the triangle with vertices $(x_i-x_0,y_i-y_0),\,i=1,\ldots,3$. The triangle with vertices $(\sqrt{a}(x_i-x_0),\sqrt{a}(y_i-y_0)),\,i=1,\ldots,3$ will have $a$ times the area of the triangle in the origin. Now move your triangle back to where it belongs to get the final answer $(\sqrt{a}(x_i-x_0)+x_0,\sqrt{a}(y_i-y_0)+y_0),\,i=1,\ldots,3$.

Edit: Let me also add that depending on what you want the little and large triangle to have incommon, the formula may change (e.g. It will be different if you want them to have a common incenter instead of a common center of mass). Well, even for a common center of mass, the formula is not unique, of course (rotations).

Answer 2

Area of a triangle is given by $S=\frac{1}{2}\begin{vmatrix}x_1-x_3\ y_1-y_3\\ x_2-x_3 \ y_2-y_3\end{vmatrix}$.
To multiply its area by $\alpha$ you can multiply each coordinate by a factor of $\sqrt{\alpha}$, i.e. $$x_i^{scaled}=\sqrt{\alpha} x_i,$$ $$y_i^{scaled}=\sqrt{\alpha} y_i.$$
Having done this, you'll most likely get the triangle moved in the direction of its center, so you'll have to compensate for difference between new and old centers to the coordinates. Triangle center of mass can be determined using these formulas: $$x_c=\frac{x_1+x_2+x_3}{3},$$ $$y_c=\frac{y_1+y_2+y_3}{3}.$$ So instead of doing $scale(triangle)$ you want to do $scale(triangle-center)+center$, or in coordinate form: $$x_i^{final}=\sqrt{\alpha}\left(x_i-x_c\right)+x_c,$$ $$y_i^{final}=\sqrt{\alpha}\left(y_i-y_c\right)+y_c.$$

Edit: oops, while I was writing this, it appeared to have been answered...