If (2x+3y)2−36(3x−2y)2=0 and 2x+3y+45=0 represents an Isosceles triangle with base angletan−16 then its area is

Given Equations (2x+3y)2−36(3x−2y)2=0and 2x+3y+45=0 comparing with lx+my+n=0∴ l=2 , m=3, n=45and let α=Tan−1(6)⇒ tanα=6Area of issoceles Δle = n2tanα(l2+m2)=16(5)6(2+3)⇒ 83

If (2x+3y)2−36(3x−2y)2=0 and 2x+3y+45=0 represents an Isosceles triangle with base angletan−16 then its area is

Given Equations (2x+3y)2−36(3x−2y)2=0and 2x+3y+45=0 comparing with lx+my+n=0∴ l=2 , m=3, n=45and let α=Tan−1(6)⇒ tanα=6Area of issoceles Δle = n2tanα(l2+m2)=16(5)6(2+3)⇒ 83

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$\begin{array}{l} I f {(\sqrt{2} x + \sqrt{3} y)}^{2} - 36 {(\sqrt{3} x - \sqrt{2} y)}^{2} = 0 a n d \sqrt{2} x + \sqrt{3} y + 4 \sqrt{5} = 0 r e p r e s e n t s a n I s o s c e l e s t r i a n g l e w i t h b a s e a n g l e \\ \tan^{- 1} 6 t h e n i t s a r e a i s \end{array}$

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$4$

$\frac{8}{3}$

$8 \sqrt{6}$

$9$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n s {(\sqrt{2} x + \sqrt{3} y)}^{2} - 36 {(\sqrt{3} x - \sqrt{2} y)}^{2} = 0 \\ a n d \sqrt{2} x + \sqrt{3} y + 4 \sqrt{5} = 0 c o m p a r i n g w i t h l x + m y + n = 0 \\ ∴ l = \sqrt{2}, m = \sqrt{3}, n = 4 \sqrt{5} \\ a n d l e t α = T a n^{- 1} (6) \\ \Rightarrow \tan α = 6 \\ A r e a o f i s s o c e l e s Δ^{l e} = \frac{n^{2}}{\tan α (l^{2} + m^{2})} \\ = \frac{16 (5)}{6 (2 + 3)} \Rightarrow \frac{8}{3} \end{array}$