Question

Find the equations of the sides of a square whose each side is of length 4 units and centre is $(1,1)$. Given that one pair of sides is parallel to $3x-4y=0$.

• A. $3x-4y+11=0,3x-4y-9=0,4x+3y+3=0,4x+3y-17=0$
• B. $3x-4y-15=0,3x-4y+5=0,4x+3y+3=0,4x+3y-17=0$
• C. $3x-4y+11=0,3x-4y-9=0,4x+3y+2=0,4x+3y-18=0$
• D. None of the above

Answer 1

The correct option is A $3x-4y+11=0,3x-4y-9=0,4x+3y+3=0,4x+3y-17=0$

Given the center of the square is A$(1,1)$ and length of the square is $4$

one side of the square is parallel to line $3x-4y=0$ hence

The equation of one side is$3x-4y+\lambda =0$

The perpendicular distance from the centre of the square to all sides is $2$

Hence $\left|\frac{3-4+\lambda }{\sqrt{3^2+4^2}}\right|=2$

$\frac{-1+\lambda }{\sqrt{25}}=\pm 2$

$-1+\lambda =\pm 10$

$\lambda =11$ and $\lambda =-9$

Hence Eq of sides $3x-4y+11=0$ and $3x-4y-9=0$

Now the other two sides of the square should be perpendicular

Hence equation of sides is $4x+3y+\mu =0$

The perpendicular distance from centre of square to all sides is $2$

Hence $\left|\frac{4+3+\mu }{\sqrt{3^2+4^2}}\right|=2$

$\frac{7+\mu }{\sqrt{25}}=\pm 2$

$7+\mu =\pm 10$

$\mu =3$ and $\mu =-17$

Hence Eq of other two sides $4x+3y+3=0$ and $4x+3y-17=0$