Question

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

• A. Square but not rhombus
• B. Rhombus
• C. Paralleogram
• D. Rectangle but not a square

Answer 1

The correct option is C Paralleogram

Given pairs

$x^2-3xy+2y^2=0$

$(x-2y)(x-y)=0$

$x-2y=0$ and $x-y=0$ are two sides

$x^2-3xy+2y^2+x-2=0$

$x=\frac{-(1-3y)\pm \sqrt{1+9y^2-6y-4(2y^2-2)}}{2}$

$x=\frac{-(1-3y)\pm \sqrt{1+9y^2-6y-8y^2+8)}}{2}$

$2x=-1+3y\pm \sqrt{y^2-6y+9)}$

$2x=-1+3y\pm \sqrt{(y-3{)}^2)}$

$2x=-1+3y\pm (y-3)$

$2x=-1+3y+y-3$ and $2x=-1+3y-y+3$

$2x-4y=-4$ and $2x-2y=2$

$x-2y=-2$ and $x-y=1$ are other two sides

$x-2y=0$ and $x-y=0$ and $x-2y=-2$ and $x-y=1$ are four sides

Here two opposite sides are parallel

Angle between $x-2y=0$ and $x-y=0$ by formula is

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\right|$

$\tan \theta =\frac{1}{3}$

Here the sides are not perpendicular

Hence it is parallelogram