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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x-$ axis. If equation of $RQ$ is $x -2y = 2$ and $PQ$ is parallel to the $x-$ axis, then the centroid of $\Delta PQR$ lies on the line

If $P = (1, 0) ; Q = (-1, 0) \,\,ane\,\, R = (2, 0)$ are three given points, then the locus of the points $S$ satisfying the relation, $SQ^2 + SR^2 = 2 SP^2$ is :

A straight the through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q.$ If $O$ is the origin and the rectangle $OPRQ$ is completed, then the locus of $R$ is:

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :