Question

If $(–4,3)$ and$(4,3)$ are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer 1

Step 1: Solve for the $x-$coordinate of third vertex:

Given two vertices of an equilateral triangle are $A(–4,3)$ and $B(4,3)$

Let the third vertex be$C(x,y)$

We know that the distance between two points $\left(x_1,y_1\right),\left(x_2,y_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Therefore,

Distance between $(x,y),(4,3)$is $=\sqrt{{\left(x-4\right)}^2+{\left(y-3\right)}^2}$ …$\left[1\right]$

Distance between $(x,y),(-4,3)$is $=\sqrt{{\left(x+4\right)}^2+{\left(y-3\right)}^2}$ …$\left[2\right]$

Distance between $(4,3),(-4,3)$is $\sqrt{{\left(4+4\right)}^2+{\left(3-3\right)}^2}=\sqrt{{\left(8\right)}^2}=8$

Given that the triangle is equilateral$\Rightarrow AB=BC=CA$

Consider $AC=BC$

$\sqrt{{\left(x+4\right)}^2+{\left(y-3\right)}^2}$ $=\sqrt{{\left(x-4\right)}^2+{\left(y-3\right)}^2}$

Squaring on both sides, we get,

${\left(x+4\right)}^2+{\left(y-3\right)}^2$ $={\left(x-4\right)}^2+{\left(y-3\right)}^2$

$$\begin{gathered} \Rightarrow {(x-4)}^2={(x+4)}^2 \\ \Rightarrow x^2-8x+16=x^2+8x+16 \\ \Rightarrow 16x=0 \\ \Rightarrow x=0 \end{gathered}$$

Step 2: Solve for the y-coordinate of third vertex:

Consider $BC=AB$

$\sqrt{{\left(x-4\right)}^2+{\left(y-3\right)}^2}=8$

$\Rightarrow$${\left(x-4\right)}^2+{(y-3)}^2=64$ …$\left[3\right]$

Substituting the value of $x$ in $\left[3\right]$

$$\begin{gathered} \Rightarrow {(0-4)}^2+{(y-3)}^2=64 \\ \Rightarrow {(y-3)}^2=64-16 \\ \Rightarrow {(y-3)}^2=48 \\ \Rightarrow \left(y-3\right)=\pm 4\sqrt{3} \\ \Rightarrow y=3+4\sqrt{3},3-4\sqrt{3} \end{gathered}$$

Consider $y=3-4\sqrt{3}$because it is given that origin is interior of triangle.

Hence the co-ordinates of the third vertex is $\left(0,3-4\sqrt{3}\right)$.