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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $(a^{2} + b^{2}) - 4 a b = 138$
$O A = O B$
$\Rightarrow a^{2} + 11^{2} = b^{2} + 37^{2} \Rightarrow a^{2} - b^{2} = 37^{2} - 11^{2} \dots (1)$

$A B = O A$
$\Rightarrow (a - b)^{2} + 26^{2} = a^{2} + 11^{2} \Rightarrow b^{2} - 2 a b = 11^{2} - 26^{2} \dots (2)$

On adding $(1) & 2 \times (2)$ , we get
$a^{2} + b^{2} - 4 a b = 37^{2} - 26^{2} + 11^{2} - 26^{2} \Rightarrow (a^{2} + b^{2}) - 4 a b = 138$

Suggest Corrections

Similar questions

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 |

\frac{a}{b}

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

\frac{a^{2}}{(a^{2} + b^{2})}

\frac{b^{2}}{(a^{2} + b^{2})}

Join BYJU'S Learning Program