Chapter 9.5, Problem 27PSC

To determine

To Calculate: Possible coordinates of the third vertex.

Answer to Problem 27PSC

The possible coordinates of the third vertex are $\left(4-\sqrt{6},3-\sqrt{6}\right)$

Explanation of Solution

Given information:

Two of the vertices of an equilateral triangle are $\left(2,1\right)$ and $\left(6,5\right)$

Formula used:

Here, we are using distance formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

  

Here, AB = AC = BC (equilateral triangle)

Here, the formula is

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $16+16={\left(y-1\right)}^2+{\left(x-2\right)}^2$

  $32=x^2+y^2-4x-2y+5$ ............. $\left(1\right)$

And, ${\left(y-5\right)}^2+{\left(x-6\right)}^2=32$

  $y^2+25-10y+x^2+36-12x=32$ .............. $\left(2\right)$

Hence, equate equation $\left(1\right)$ and $\left(2\right)$

So, $x^2+y^2-4x-2y+5=x^2+y^2-12x-10y+61$

  $8x+8y=56$

  $x+y=7$

  $x=7-y$ (put in equation $\left(1\right)$ )

So, $32=\left(7-y^2\right)+y^2-4\left(7-y\right)-2y+5$

  $32=49+2y^2-12y+5-28$

  $2y^2-12y=6$

So, $y=\frac{6\pm \sqrt{36-12}}{2}$

  $y=\frac{6\pm \sqrt{24}}{2}$

  $y=3\pm \sqrt{6}$

And, $x=4\pm \sqrt{6}$

Conclusion: Hence,the possible coordinates of the third vertex are $\left(4-\sqrt{6},3-\sqrt{6}\right)$