The triangle formed by {x^2} - 9{y^2} = 0 and | vedclass

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(A) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.We have ${x^2} - 9{y^2} = 0$,which can be factored as $(x - 3y)(x + 3y) = 0$.So the equations of

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Isosceles

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(A) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.We have ${x^2} - 9{y^2} = 0$,which can be factored as $(x - 3y)(x + 3y) = 0$.So the equations of the lines are:$x - 3y = 0$ ..... $(i)$$x + 3y = 0$ ..... $(ii)$$x = 4$ ..... $(iii)$By solving these equations,we find the vertices of the triangle:Intersection of $(i)$ and $(ii)$ is $A(0, 0)$.Intersection of $(i)$ and $(iii)$ is $C(4, 4/3)$.Intersection of $(ii)$ and $(iii)$ is $B(4, -4/3)$.Now,we calculate the lengths of the sides:$AB = \sqrt{(4 - 0)^2 + (-4/3 - 0)^2} = \sqrt{16 + 16/9} = \sqrt{160/9} = \frac{4\sqrt{10}}{3}$$AC = \sqrt{(4 - 0)^2 + (4/3 - 0)^2} = \sqrt{16 + 16/9} = \sqrt{160/9} = \frac{4\sqrt{10}}{3}$$BC = \sqrt{(4 - 4)^2 + (4/3 - (-4/3))^2} = \sqrt{0 + (8/3)^2} = 8/3$Since $AB = AC$,the triangle $ABC$ is an isosceles triangle.

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is

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