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9.Straight Line
easy
The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is
A
Isosceles
B
Equilateral
C
Right angled
D
None of these
Solution
(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.
Standard 11
Mathematics
