4 if two vertices of an equilateral triangle be (0, 0); (3, sqrt(3)) find the third vertex.

Text Solution

Solution : Let A=`(3,sqrt(3))`<br> and O be any point then, <br> `O A^{2}=(3-0)^{2}+(sqrt{3}-0)^{2}=12`,<br><br> `O B^{2}=x^{2}+y^{2}` and,`A B^{2}=(x-3)^{2}+(y-sqrt{3})^{2}`<br><br> `Rightarrow A B^{2}=x^{2}+y^{2}-6 x-2 y+12`<br><br> `therefore O A^{2}=O B^{2}=A B^{2}`<br><br> `Rightarrow O A^{2}=O B^{2}` and `O B^{2}=A B^{2}`<br><br> `Rightarrow x^{2}+y^{2}=12`<br><br> and,`x^{2}+y^{2}=x^{2}+y^{2}-6 x-2 sqrt{3} y+12`<br><br> `Rightarrow x^{2}+y^{2}=12` and `6 x+2 sqrt{3} y=12`<br><br> `Rightarrow x^{2}+y^{2}=12` and `3 x+sqrt{3} y=6`<br><br> `Rightarrow x^{2}+(frac{6-3 x}{sqrt{3}})^{2}=12[therefore 3 x+sqrt{3} y=6 therefore y=frac{6-3 x}{sqrt{3}}]`<br><br> `Rightarrow 3 x^{2}+(6-3 x)^{2}=36`<br><br> `Rightarrow 12 x^{2}-36 x=0 Rightarrow x=0,3`<br><br> `therefore x=0 Rightarrow sqrt{3} y=6`<br><br> `Rightarrow y=frac{6}{sqrt{3}}=2 sqrt{3}`<br><br> and `x=3 Rightarrow 9+sqrt{3} y=6`<br><br> `Rightarrow y=frac{6-9}{sqrt{3}}=-sqrt{3}`<br><br> Hence, the coordinates of the third vertex `B` are `(0,2 sqrt{3})` or `(3,-sqrt{3})`.<br><br>

Two vertices of an equilateral triangle are 0,0 and √3, √3. Find the third vertex.

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