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

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Question

(a) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.

We have ${x^2} - 9{y^2} = 0$

So equation of line is,

$x - 3y = 0$.....$(i)$

$x + 3y

Vedclass · Accepted Answer

Isosceles

Vedclass · Answer

(a) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.

We have ${x^2} - 9{y^2} = 0$

So equation of line is,

$x - 3y = 0$.....$(i)$

$x + 3y = 0$.....$(ii)$

$x = 4$.....$(iii)$

By solving $(i)$, $(ii)$ and $(iii)$ we get

$A(0,\,0),\,\,B\,\left( {4,\,\frac{{ - 4}}{3}} \right),\,\,C\,\left( {4,\,\frac{4}{3}} \right)$

Now, $AB = \sqrt {{{(4 - 0)}^2} + {{\left( {0 + \frac{4}{3}} \right)}^2}} = \frac{{4\sqrt {10} }}{3}$

$AC = \sqrt {{{(4 - 0)}^2} + {{\left( {0 - \frac{4}{3}} \right)}^2}} = \frac{{4\sqrt {10} }}{3}$

$BC = \sqrt {{{(4 - 4)}^2} + {{\left( {\frac{4}{3} + \frac{4}{3}} \right)}^2}} = \frac{8}{3}$

Hence $ABC$ is an isosceles triangle.

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

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