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help.

sangangsx  Mar 31, 2018

Best Answer 

 #1
avatar+6943 
+3

Let the coordinates of the third point be  (x, y) .  Since the triangle is equilateral...

 

distance between  (x, y)  and  (-1, 2)   =   distance between  (x, y)  and  (4, 2)

 

\(\sqrt{(-1-x)^2+(2-y)^2}\,=\,\sqrt{(4-x)^2+(2-y)^2} \\ (-1-x)^2+(2-y)^2\,=\,(4-x)^2+(2-y)^2 \\ (-1-x)^2\,=\,(4-x)^2 \\ (-1-x)\,=\,\pm(4-x)\\ \begin{array}\ (-1-x)=+(4-x)\qquad&\text{or}&\qquad(-1-x)=-(4-x)\\ -1-x=4-x&&\qquad-1-x=-4+x\\ -1=4&&\qquad-1=-4+2x\\ \text{not a solution}&&\qquad3=2x\\ &&\qquad\frac32=x \end{array}\)

 

So we know that the x coordinate must be  3/2 , which is 1.5 .

 

To find the y coordinate, we need to make another equation.

 

distance between  (-1,2)  and  (4, 2)   =   4 - -1   =   5      So...

 

distance between (-1, 2)  and  (1.5, y)   =   5

 

\(\sqrt{(-1-1.5)^2+(2-y)^2}=5\\ \sqrt{6.25+(2-y)^2}=5\\ 6.25+(2-y)^2=25\\(2-y)^2=18.75\\2-y=\pm\sqrt{18.75}\\ -y=\pm\sqrt{18.75}-2\\ y=\pm\sqrt{18.75}+2\\ \begin{array}\ y=\sqrt{18.75}+2\qquad\text{or}\qquad&&y=-\sqrt{18.75}+2\\ y\approx6.3&&y\approx-2.3 \end{array}\)

 

So the solutions for  (x, y)  are:  (1.5, 6.3)  and  (1.5, -2.3)

hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018
Sort: 

1+0 Answers

 #1
avatar+6943 
+3
Best Answer

Let the coordinates of the third point be  (x, y) .  Since the triangle is equilateral...

 

distance between  (x, y)  and  (-1, 2)   =   distance between  (x, y)  and  (4, 2)

 

\(\sqrt{(-1-x)^2+(2-y)^2}\,=\,\sqrt{(4-x)^2+(2-y)^2} \\ (-1-x)^2+(2-y)^2\,=\,(4-x)^2+(2-y)^2 \\ (-1-x)^2\,=\,(4-x)^2 \\ (-1-x)\,=\,\pm(4-x)\\ \begin{array}\ (-1-x)=+(4-x)\qquad&\text{or}&\qquad(-1-x)=-(4-x)\\ -1-x=4-x&&\qquad-1-x=-4+x\\ -1=4&&\qquad-1=-4+2x\\ \text{not a solution}&&\qquad3=2x\\ &&\qquad\frac32=x \end{array}\)

 

So we know that the x coordinate must be  3/2 , which is 1.5 .

 

To find the y coordinate, we need to make another equation.

 

distance between  (-1,2)  and  (4, 2)   =   4 - -1   =   5      So...

 

distance between (-1, 2)  and  (1.5, y)   =   5

 

\(\sqrt{(-1-1.5)^2+(2-y)^2}=5\\ \sqrt{6.25+(2-y)^2}=5\\ 6.25+(2-y)^2=25\\(2-y)^2=18.75\\2-y=\pm\sqrt{18.75}\\ -y=\pm\sqrt{18.75}-2\\ y=\pm\sqrt{18.75}+2\\ \begin{array}\ y=\sqrt{18.75}+2\qquad\text{or}\qquad&&y=-\sqrt{18.75}+2\\ y\approx6.3&&y\approx-2.3 \end{array}\)

 

So the solutions for  (x, y)  are:  (1.5, 6.3)  and  (1.5, -2.3)

hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018
edited by hectictar  Mar 31, 2018

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