The triangle formed by $2 x + 3 y + 7 = 0 a n d t h e p a i r {(2 x + 3 y)}^{2} - 7 {(3 x - 2 y)}^{2} = 0 i s$

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Isosceles

Right angled isosceles

Equilateral

Right angled

answer is A.

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Detailed Solution

Given Equation 2x + 3y – 7 = 0 –(1) and

$\begin{array}{l} {(2 x + 3 y)}^{2} - 7 {(3 x - 2 y)}^{2} = 0 \\ \Rightarrow {(2 x + 3 y)}^{2} - {(\sqrt{7} (3 x - 2 y))}^{2} = 0 \\ \Rightarrow [2 x + 3 y + \sqrt{7} (3 x - 2 y)] [(2 x + 3 y) - \sqrt{7} (3 x - 2 y)] = 0 \\ \Rightarrow x (2 + 3 \sqrt{7}) + y (3 - 2 \sqrt{7}) = 0 - - (2) \\ x (2 - 3 \sqrt{7}) + y (3 + 2 \sqrt{7}) = 0 - - (3) \end{array}$

$\begin{array}{l} A n g l e b / w (1) & (2) \\ Cos A = \frac{|2 (2 + 3 \sqrt{7}) + 3 (3 - 2 \sqrt{7})|}{\sqrt{4 + 9} \sqrt{{(2 + 3 \sqrt{7})}^{2} + \sqrt{{(3 - 2 \sqrt{7})}^{2}}}} \\ = \frac{|4 + 6 \sqrt{7} + 9 - 6 \sqrt{7}|}{\sqrt{13} \sqrt{4 + 63 + 9 + 28}} \Rightarrow \frac{13}{\sqrt{13} \sqrt{104}} = \sqrt{\frac{13}{104}} \\ A n g l e b / w (1) & (3) \cos B = \frac{|2 (2 - 3 \sqrt{7}) + 3 (3 + 2 \sqrt{7})|}{\sqrt{4 + 9} \sqrt{{(2 - 3 \sqrt{7})}^{2} + {(3 + 2 \sqrt{7})}^{2}}} \\ = \frac{|4 - 6 \sqrt{7} + 9 + 6 \sqrt{7}|}{\sqrt{13} \sqrt{4 + 63 + 9 + 28}} \Rightarrow \frac{13}{\sqrt{13} \sqrt{104}} = \sqrt{\frac{13}{104}} \end{array}$

$\begin{array}{l} A n g l e b / w (2) & (3) \cos C = \frac{|(2 + 3 \sqrt{7}) + (3 - 2 \sqrt{7}) (3 + 2 \sqrt{7})|}{\sqrt{{(2 + 3 \sqrt{7})}^{2} + {(3 - 2 \sqrt{7})}^{2}} \sqrt{{(2 - 3 \sqrt{7})}^{2} + {(3 + 2 \sqrt{7})}^{2}}} \\ = \frac{|(4 - 63) + (9 - 28)|}{\sqrt{104} \sqrt{104}} \Rightarrow \frac{78}{104} \Rightarrow \frac{39}{52} = \frac{3}{4} \\ ∴ ∠ A = ∠ B \\ Δ^{l e} i s I s o s c e l e s \end{array}$