Question

ABC is an equilateral triangle with coordinates of A(-2,0) and B(2,0). Find the coordinates of C ​

Answer 1

SOLUTION

GIVEN

ABC is an equilateral triangle with coordinates of A(-2,0) and B(2,0).

TO DETERMINE

The coordinates of C

EVALUATION

Let the coordinates of C = (a, b)

Since ordinate of both A and B is zero

AB = 2 + 2 = 4 unit

Since ABC is an equilateral triangle

AB = BC = CA

Now BC = CA gives

$\sf{ \sqrt{ {(a + 2)}^{2} + {b}^{2} } = \sqrt{ {(a - 2)}^{2} + {b}^{2} } }$

$\sf{ \implies \: {(a + 2)}^{2} + {b}^{2} = {(a - 2)}^{2} + {b}^{2} }$

$\sf{ \implies \: {(a + 2)}^{2} = {(a - 2)}^{2} }$

$\sf{ \implies \: 8 a = 0}$

$\sf{ \implies \: a = 0}$

Again BC = CA gives

$\sf{ \sqrt{ {(a - 2)}^{2} + {b}^{2} } = 4}$

$\sf{ \implies \:{(a - 2)}^{2} + {b}^{2} = {4}^{2} }$

$\sf{ \implies \:{(0 - 2)}^{2} + {b}^{2} =16}$

$\sf{ \implies \: {b}^{2} =16 - 4}$

$\sf{ \implies \: {b}^{2} =12}$

$\sf{ \implies \: b = \pm \: 2 \sqrt{3} }$

Hence the required coordinates of C is

( 0 , 2√3 ) or ( 0 , - 2√3 )

FINAL ANSWER

The coordinates of C is

( 0 , 2√3 ) or ( 0 , - 2√3 )

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Answer 2

Solution :-

Let us assume that, the coordinates of C are (x, y) .

since ∆ABC is an equilateral triangle . All sides will be of equal length .

So,

→ AB = BC = AC

then,

→ AB = √{(-2 - 2)² + (0 - 0)²} = √(-4)² = √16 = 4 units .

we gets ,

→ AB = BC = AC = 4 units .

now,

→ Mid point of AB are = (-2 + 2)/2 and (0 + 0)/2 = (0, 0)

therefore, we can conclude that, C lies on y - axis .

So,

→ AO = 4/2 = 2 units { Perpendicular height bisects the base in equaliteral ∆. }

using pythagoras now we get,

→ CO = √(AC² - AO²)

→ CO = √(4² - 2²)

→ CO = √(16 - 4)

→ CO = √(12)

→ CO = ± 2√3

hence, the coordinates of C are (0, 2√3) or (0 , - 2√3) .

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