Question

If $\left( {0,0} \right)$, $\left( {3,0} \right)$ and $\left( {x,y} \right)$ are the vertices of an equilateral then the value of $x$ and $y$ is
A) $\left( {\dfrac{3}{2},\dfrac{{3\sqrt 3 }}{2}} \right)$or $\left( {\dfrac{3}{2},\dfrac{{ - 3\sqrt 3 }}{2}} \right)$
B) \[\left( {\dfrac{1}{2},\sqrt 2 } \right)\]or \[\left( {\dfrac{1}{2}, - \sqrt 2 } \right)\]
C) \[\left( {\dfrac{1}{3},1} \right)\] or \[\left( {\dfrac{1}{3}, - 1} \right)\]
D) None of these

Answer 1

Hint: In order to find the value of $x$ and $y$, use the property of an Equilateral triangle, that is for an equilateral triangle all the sides are equal. So, if the vertices are given then the distance between the vertices for one side would be equal to the distance of the vertices of other sides. So, just place the vertex point on the distance formula and get the results.

Complete step by step solution:
We are given three vertices $\left( {0,0} \right)$, $\left( {3,0} \right)$ and $\left( {x,y} \right)$.
We are going to find the distance between two vertices once at a time.
Taking the first two vertices $\left( {0,0} \right)$, $\left( {3,0} \right)$.
From the distance formula, we know that the distance between two points for $\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ is:
$D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Comparing $\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ with $\left( {0,0} \right)$, $\left( {3,0} \right)$, we get $\left( {{x_1} = 0,{y_1} = 0} \right)$ and $\left( {{x_2} = 3,{y_2} = 0} \right)$.
To find the distance between the two vertices, substituting this value in the distance formula and we get:
$
  D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \\
  D = \sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} \\
 $
Solving the inner brackets and the square root for the above equation, and we get:
$
  D = \sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} \\
\Rightarrow D = \sqrt {{{\left( 3 \right)}^2} + {{\left( 0 \right)}^2}} \\
\Rightarrow D = \sqrt 9 \\
\Rightarrow D = \pm 3 \\
 $
Similarly, substituting the next two vertices $\left( {0,0} \right)$ and $\left( {x,y} \right)$ in the distance formula, and we get:
$
  D = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} \\
\Rightarrow D = \sqrt {{{\left( x \right)}^2} + {{\left( y \right)}^2}} \\
\Rightarrow D = \sqrt {{x^2} + {y^2}} \\
 $
Squaring both the sides of the above equation:
$
  D = \sqrt {{x^2} + {y^2}} \\
\Rightarrow {\left( D \right)^2} = {\left( {\sqrt {{x^2} + {y^2}} } \right)^2} \\
\Rightarrow {D^2} = {x^2} + {y^2} \\
 $
But since then, we got that value of $D = \pm 3$. So, substituting this vale above and solving it:
$
  {D^2} = {x^2} + {y^2} \\
\Rightarrow {\left( { \pm 3} \right)^2} = {x^2} + {y^2} \\
\Rightarrow 9 = {x^2} + {y^2} \\
\Rightarrow {x^2} + {y^2} = 9 \\
 $
Marking the equation, ${x^2} + {y^2} = 9$ as (1).
Now, taking the other left vertices that is $\left( {3,0} \right)$ and $\left( {x,y} \right)$. Substituting in the distance formula, and we get:
$
  D = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 0} \right)}^2}} \\
\Rightarrow D = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( y \right)}^2}} \\
 $
Opening the inner bracket by using the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, and solving it further and we get:
$
  D = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 0} \right)}^2}} \\
\Rightarrow D = \sqrt {{x^2} + 9 + 6x + {{\left( y \right)}^2}} \\
\Rightarrow D = \sqrt {{x^2} + {y^2} + 6x + 9} \\
 $
Squaring both the sides of the above equation:
$
  D = \sqrt {{x^2} + {y^2} + 6x + 9} \\
\Rightarrow {\left( D \right)^2} = {\left( {\sqrt {{x^2} + {y^2} + 6x + 9} } \right)^2} \\
\Rightarrow {D^2} = {x^2} + {y^2} + 6x + 9 \\
 $
But since then, we got the value of $D = \pm 3$. So, substituting this value above and solving it:
$
  {D^2} = {x^2} + {y^2} + 6x + 9 \\
\Rightarrow {\left( { \pm 3} \right)^2} = {x^2} + {y^2} + 6x + 9 \\
\Rightarrow 9 = {x^2} + {y^2} + 6x + 9 \\
 $
Subtracting the side by $9$, and we get:
$
  9 = {x^2} + {y^2} + 6x + 9 \\
\Rightarrow 9 - 9 = {x^2} + {y^2} + 6x + 9 - 9 \\
\Rightarrow {x^2} + {y^2} + 6x = 0 \\
 $
Marking the equation, ${x^2} + {y^2} + 6x = 0$ as (2).
Subtracting (1) from (2) and, we get:
$
  \left( {{x^2} + {y^2} + 6x} \right) - \left( {{x^2} + {y^2} - 9} \right) = 0 \\
\Rightarrow {x^2} + {y^2} + 6x - {x^2} - {y^2} - 9 = 0 \\
\Rightarrow 6x - 9 = 0 \\
 $
Adding both the side by $9$ and dividing by $6$ to get the value of $x$:
\[
  6x - 9 = 0 \\
\Rightarrow 6x - 9 + 9 = 0 + 9 \\
\Rightarrow \dfrac{{6x}}{6} = \dfrac{9}{6} \\
\Rightarrow x = \dfrac{3}{2} \\
 \]
Since, we got the value of $x$, now we need to find the value of $y$, for this substituting this value in equation (1):
$
  {x^2} + {y^2} = 9 \\
\Rightarrow {\left( {\dfrac{3}{2}} \right)^2} + {y^2} = 9 \\
\Rightarrow \dfrac{9}{4} + {y^2} = 9 \\
 $
Subtracting both the sides by $\dfrac{9}{4}$, and we get:
$
  \dfrac{9}{4} + {y^2} = 9 \\
\Rightarrow \dfrac{9}{4} + {y^2} - \dfrac{9}{4} = 9 - \dfrac{9}{4} \\
\Rightarrow {y^2} = \dfrac{{36 - 9}}{4} = \dfrac{{27}}{4} \\
\Rightarrow {y^2} = \dfrac{{27}}{4} \\
 $
Since, we know that if ${y^2} = 9$, then $y = \pm \sqrt 9 = \pm 3$. So, using this method in the above equation to find the value of $y$, and we get:
\[
  {y^2} = \dfrac{{27}}{4} \\
\Rightarrow y = \sqrt {\dfrac{{27}}{4}} = \pm \dfrac{{3\sqrt 3 }}{2} \\
\Rightarrow \left( {y = \dfrac{{3\sqrt 3 }}{2}} \right)or\left( {y = \dfrac{{ - 3\sqrt 3 }}{2}} \right) \\
 \]
According to these, the values of \[x\]and \[y\] are \[\left( {\dfrac{3}{2},\dfrac{{3\sqrt 3 }}{2}} \right)\] or \[\left( {\dfrac{3}{2},\dfrac{{ - 3\sqrt 3 }}{2}} \right)\], which matches with the option A.

Therefore, the correct option is Option (A).

Note:
> Always prefer to go step by step, by taking two vertices at a time for ease otherwise it may lead to error.
> Do not commit a mistake by taking only the positive value, because the points in the cartesian plane have both the negative and positive points.