The area of the triangle formed by the lines | vedclass

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(B) The given equation $x^2 - 4y^2 = 0$ can be factored as $(x - 2y)(x + 2y) = 0$,which represents two lines: $x - 2y = 0$ and $x + 2y = 0$.These line

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$\frac{a^2}{2}$

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(B) The given equation $x^2 - 4y^2 = 0$ can be factored as $(x - 2y)(x + 2y) = 0$,which represents two lines: $x - 2y = 0$ and $x + 2y = 0$.These lines intersect at the origin $A(0, 0)$.The third line is $x = a$.To find the vertices of the triangle,we find the intersection points:$1$. Intersection of $x - 2y = 0$ and $x = a$: $a - 2y = 0 \implies y = \frac{a}{2}$. So,$C = (a, \frac{a}{2})$.$2$. Intersection of $x + 2y = 0$ and $x = a$: $a + 2y = 0 \implies y = -\frac{a}{2}$. So,$B = (a, -\frac{a}{2})$.$3$. Intersection of $x - 2y = 0$ and $x + 2y = 0$ is $A = (0, 0)$.The base of the triangle is the vertical segment $BC$ on the line $x = a$. The length of the base is $|\frac{a}{2} - (- \frac{a}{2})| = |a| = a$ (assuming $a > 0$).The height of the triangle from vertex $A$ to the base $BC$ is the horizontal distance from $x = 0$ to $x = a$,which is $a$.Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes a 	imes a = \frac{a^2}{2}$.

The area of the triangle formed by the lines $x^2 - 4y^2 = 0$ and $x = a$ is

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