MCQ
The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is
- ✓Isosceles
- BEquilateral
- CRight angled
- DNone of these
✓
Answer
Correct option: A.
Isosceles
a
(a) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.
(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.
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