Question

Type of quadrilateral formed by the two pairs of lines $6x^2-5xy-6y^2=0$ and $6x^2-5xy-6y^2+x+5y-1=0$ is

• A. square
• B. rhombus
• C. parallelogram
• D. rectangle

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A square
B rhombus
C parallelogram
D rectangle

$6x^2-5xy-6y^2=0$
$x=\frac{5y\pm \sqrt{25y^2+144y^2}}{12}$
$12x=5y\pm 13y$
$12x=18y$ and $12x=-8y$
$\to 2x-3y=0,3x+2y=0$
These lines are mutually perpendicular. Since these are adjacent sides, hence the angle included within the sides is ${90}^0$ ...(i)
$6x^2+x(-5y+1)-6y^2+5y-1=0$
$x=\frac{5y-1\pm \sqrt{25y^2+1-10y-24(-6y^2+5y-1)}}{12}$
$12x=5y-1\pm \left(\sqrt{169y^2-130y+25}\right)$
$12x=5y-1\pm (13y-5)$
$12x=18y-6$ and $12x=-8y+4$
Thus
$2x-3y=-1$ and $3x+2y=1$
Again these two lines are perpendicular to another. Thus the angle between then is ${90}^0$.
Now one pair of opposite sides are
$2x-3y=-1$ and $2x-3y=0$ distance between the lines is 1 unit.
Another pair of opposite sides are
$3x+2y=1$ and $3x+2y=0$ ..distance between the lines is 1 unit.
Thus the adjacent sides are mutually parallel and distance between each opposite side is one unit.
Hence it is a square.