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(B) The given pair of lines is $23x^2 - 48xy + 3y^2 = 0$.The area of the triangle formed by the pair of lines $ax^2 + 2hxy + by^2 = 0$ and the line $l

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$\frac{25}{13 \sqrt{3}}$

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(B) The given pair of lines is $23x^2 - 48xy + 3y^2 = 0$.The area of the triangle formed by the pair of lines $ax^2 + 2hxy + by^2 = 0$ and the line $lx + my + n = 0$ is given by the formula:$	ext{Area} = \frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|}$Here,$a = 23$,$h = -24$,$b = 3$,$l = 2$,$m = 3$,and $n = 5$.First,calculate $\sqrt{h^2 - ab} = \sqrt{(-24)^2 - (23)(3)} = \sqrt{576 - 69} = \sqrt{507} = 13 \sqrt{3}$.Next,calculate the denominator $|am^2 - 2hlm + bl^2| = |23(3)^2 - 2(-24)(2)(3) + 3(2)^2| = |23(9) + 144(2) + 3(4)| = |207 + 288 + 12| = |507| = 507$.Now,substitute the values into the area formula:$	ext{Area} = \frac{5^2 	imes 13 \sqrt{3}}{507} = \frac{25 	imes 13 \sqrt{3}}{507} = \frac{25 	imes 13 \sqrt{3}}{39 	imes 13} = \frac{25}{39} \sqrt{3} = \frac{25}{13 \sqrt{3}} 	ext{ sq. units}$.

The area of the triangle formed by the pair of lines $23x^2 - 48xy + 3y^2 = 0$ with the line $2x + 3y + 5 = 0$ is

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