Let $O = (0, 0), A = (a, 11), B = (b, 37)$ are the vertices of an equilateral triangle $O A B$ , then $a$ and $b$ satisfy the relation

(a^{2} + b^{2}) - 4 a b = 138

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(a^{2} + b^{2}) - a b = 124

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(a^{2} + b^{2}) + 3 a b = 130

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(a^{2} + b^{2}) - 3 a b = 138

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Solution

The correct option is B $(a^{2} + b^{2}) - 4 a b = 138$
Let $C$ is the middle point of side $A B$
$∴$ Coordinate of $C \equiv (\frac{a + b}{2}, \frac{11 + 37}{2})$
$C \equiv (\frac{a + b}{2}, 24)$
$∵$ $O A B$ is an equilateral triangle
$∴ O C = \sqrt{3} C A \Rightarrow {(O C)}^{2} = {(\sqrt{3} C A)}^{2}$ (By squaring both side)
$\Rightarrow {(O C)}^{2} = 3 {(C A)}^{2}$
$\Rightarrow {(\frac{a + b}{2} - 0)}^{2} + {(24 - 0)}^{2} = 3 [{(a - \frac{a + b}{2})}^{2} + {(11 - 24)}^{2}]$
$\Rightarrow \frac{a^{2} + b^{2} + 2 a b}{4} + 576 = 3 [{(\frac{a - b}{2} - 0)}^{2} + {(- 13)}^{2}] \Rightarrow a^{2} + b^{2} - 4 a b = 138$

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Similar questions

O = (0, 0), A = (a, 11)

B = (b, 37)

O A B

a

b

Simplify $\frac{a - b}{a} + \frac{a + b}{b}$

State True or False:

On subtracting $a^{2} + a b + b^{2}$ from $4 a^{2} - 3 a b + 2 b^{2}$ , the answer is $3 a^{2} - 4 a b + b^{2}$ .

(a \neq b)

| a^{2} - b^{2} | = | a^{2} + b^{2} - 2 a b |

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