Question

The four straight lines given by the equations $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ lie along the sides of a

• A. Square
• B. Rhombus
• C. Rectangle
• D. None of these

Answer 1

The correct option is A Square

taking $y$ is constant and finding the value of$x$ by roots formula.

$12x^2+\left(7y\right)x-12y^2=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{\left(7y\right)-\sqrt{(7y{)}^2-4(-12)y^{2}12}}{2\times 12}$

$\frac{\Rightarrow x=-7y\pm \sqrt{49y^2+576y^2}}{2\times 12}$

$\Rightarrow 24x=-7y\pm \sqrt{625}y$

$\Rightarrow 24xy=-7y\pm 25y$.........(1)

Again $12x^2+(7y-1)x+(-12y^2-1+7y)=0$

$\therefore x=\frac{-(7y-1)\pm \sqrt{(7y-1{)}^2-4}\left(12\right)(7y-1-12y^2)}{24}$

$\Rightarrow x=\frac{1-7y\pm \sqrt{49y^2+1-14y-336y+48+576y^2}}{24}$

$\Rightarrow 24x=1-7y\pm \sqrt{625y^2-350y+49}$

$\Rightarrow 24x=1-7y\pm (25y-7)$

$\Rightarrow x=\frac{1-7y\pm 25y-7}{24}$........(ii)

from (i) and (ii) we can clearly see co.efficient of $x$ and $y$ are same so slope are sample $m_1=\frac{24}{18},m_2=-\frac{24}{32}$

So $m_1{m}_2=-1$

$\therefore$ it is a square