Question

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

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

Answer 1

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

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

$6x^2-5xy-6y^2=0$ represents a pair of perpendicular straight lines through the origin (as $a+b=0$) and

$6x^2-5xy-6y^2+x+5y-1=0$ represents a pair of perpendicular straight lines parallel to the above straight lines.
Let $OABC$ be the rectangle formed by the two pair of lines given

$6x^2-5xy-6y^2+x+5y-1-(6x^2-5xy-6y^2)=0$ i.e., $x+5y-1=0$..............( i)

Since, $6x^2-5xy-6y^2=0\Rightarrow (2x-3y)(3x+2y)=0$

Let the equation of $OA$ be $2x-3y=0$............(ii)

and of $OC$ be $3x+2y=0$...............(iii)

Solving (i) and (ii) we get coordinates of $A\equiv (\frac{3}{13},\frac{2}{13})$ and solving (i) and (iii) we get coordinates of $C\equiv (-\frac{2}{13},\frac{3}{13})$

$A\equiv (\frac{3}{13},\frac{2}{13})$ so that $OABC$ is a square. And a square is special case of a rhombus, a parallelogram and a rectangle.