# If the three points (0, 1), (0, -1) and (x, 0) are the vertices of an equilateral triangle, then the value x are:

$

A.{\text{ }}\sqrt 3 ,\sqrt 2 \\

B.{\text{ }}\sqrt 3 , - \sqrt 3 \\

C.{\text{ }} - \sqrt 5 ,\sqrt 3 \\

D.{\text{ }}\sqrt 2 , - \sqrt 2 \\

$

Answer

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Hint: In this question apply the concept that in equilateral triangle all the sides of equilateral triangle is same and later on apply the distance formula between two points, so use these concepts to reach the solution of the question.

The given vertices of the equilateral triangle is (0, 1), (0, -1) and (x, 0)

Now as we know all that in an equilateral triangle all the sides are equal.

So, let the vertices of the triangle be A, B, C.

Therefore,

A = (0, 1), B = (0, -1), C = (x, 0)

So, according to condition of equilateral triangle we have,

AB = BC = CA

So first calculate the distance AB according to distance formula between two points which is given as

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

Now let's consider

A = (0, 1) $ \equiv \left( {{x_1},{y_1}} \right)$, B = (0, -1) $ \equiv \left( {{x_2},{y_2}} \right)$, C = (x, 0) $ \equiv \left( {{x_3},{y_3}} \right)$

Therefore AB = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} = \sqrt {{{\left( { - 2} \right)}^2}} = 2$

Therefore BC = $\sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} + {{\left( {{y_3} - {y_2}} \right)}^2}} = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {0 - \left( { - 1} \right)} \right)}^2}} = \sqrt {{x^2} + {{\left( 1 \right)}^2}} = \sqrt {{x^2} + 1} $

Therefore CA = $\sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2}} = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = \sqrt {{x^2} + {{\left( { - 1} \right)}^2}} = \sqrt {{x^2} + 1} $

Now as we know all the sides are equal therefore

AB = BC = CA

$ \Rightarrow 2 = \sqrt {{x^2} + 1} $

Now squaring both sides we have,

$

\Rightarrow {2^2} = {x^2} + 1 \\

\Rightarrow 4 - 1 = {x^2} \\

\Rightarrow {x^2} = 3 \\

$

Now take square root we have,

$ \Rightarrow x = \pm \sqrt 3 $

$ \Rightarrow x = \sqrt 3 , - \sqrt 3 $

So, the required value of x is $\sqrt 3 , - \sqrt 3 $.

Hence, option (b) is correct.

Note: In such types of questions the key concept we have to remember is that always recall the distance formula between two points which is stated above then we all know that all the sides of the equilateral triangle is equal, so calculate the distances and equate them as above and simplify, we will get the required value of x, which is the required answer.

The given vertices of the equilateral triangle is (0, 1), (0, -1) and (x, 0)

Now as we know all that in an equilateral triangle all the sides are equal.

So, let the vertices of the triangle be A, B, C.

Therefore,

A = (0, 1), B = (0, -1), C = (x, 0)

So, according to condition of equilateral triangle we have,

AB = BC = CA

So first calculate the distance AB according to distance formula between two points which is given as

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

Now let's consider

A = (0, 1) $ \equiv \left( {{x_1},{y_1}} \right)$, B = (0, -1) $ \equiv \left( {{x_2},{y_2}} \right)$, C = (x, 0) $ \equiv \left( {{x_3},{y_3}} \right)$

Therefore AB = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} = \sqrt {{{\left( { - 2} \right)}^2}} = 2$

Therefore BC = $\sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} + {{\left( {{y_3} - {y_2}} \right)}^2}} = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {0 - \left( { - 1} \right)} \right)}^2}} = \sqrt {{x^2} + {{\left( 1 \right)}^2}} = \sqrt {{x^2} + 1} $

Therefore CA = $\sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2}} = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {0 - 1} \right)}^2}} = \sqrt {{x^2} + {{\left( { - 1} \right)}^2}} = \sqrt {{x^2} + 1} $

Now as we know all the sides are equal therefore

AB = BC = CA

$ \Rightarrow 2 = \sqrt {{x^2} + 1} $

Now squaring both sides we have,

$

\Rightarrow {2^2} = {x^2} + 1 \\

\Rightarrow 4 - 1 = {x^2} \\

\Rightarrow {x^2} = 3 \\

$

Now take square root we have,

$ \Rightarrow x = \pm \sqrt 3 $

$ \Rightarrow x = \sqrt 3 , - \sqrt 3 $

So, the required value of x is $\sqrt 3 , - \sqrt 3 $.

Hence, option (b) is correct.

Note: In such types of questions the key concept we have to remember is that always recall the distance formula between two points which is stated above then we all know that all the sides of the equilateral triangle is equal, so calculate the distances and equate them as above and simplify, we will get the required value of x, which is the required answer.

Last updated date: 02nd Oct 2023

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