Chapter E, Problem 13.12EX

To determine

To find the third vertices of equilateral triangle, whose other vertices are $(r,0)$ and $(-r,0)$ .

Answer to Problem 13.12EX

Option (b)

Explanation of Solution

Given information: Vertices of the triangle are: $(r,0)$ and $(-r,0)$ .

Concept used:

The distance between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is

  $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let ABC be the triangle with vertices as $A(r,0),B(-r,0)$ and $C(a,b)$ .

Using distance formula,

  $\begin{array}{l}AB=\sqrt{4r^2}=2r;\\ BC=\sqrt{{(a+r)}^2+b^2};\\ AC=\sqrt{{(a-r)}^2+b^2}\end{array}$

For an equilateral triangle, AB = BC = AC

  $\begin{array}{l}{(}^a+b^2={(}^a+b^2\\ a^2+r^2+2ar+b^2=a^2+r^2-2ar+b^2\\ 4ar=0\\ \Rightarrow a=0\end{array}$

Also,

  $\begin{array}{l}\sqrt{r^2+b^2}=\sqrt{4r^2}\\ r^2+b^2=4r^2\\ 3r^2=b^2\\ b=r\sqrt{3}\end{array}$

Hence, the third coordinate of the equilateral triangle is $(0,r\sqrt{3})$ .