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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 $3x-4y=0$ .
Option: 1
$3x-4y+11=0,3x-4y-9=0,4x+3y+3=0,4x+3y-17=0$

Option: 2
$3x-4y-15=0,3x-4y+5=0,4x+3y+3=0,4x+3y-17=0$

Option: 3
$3x-4y+11=0,3x-4y-9=0,4x+3y+2=0,4x+3y-18=0$

Option: 4
$None$

Answers (1)

best_answer

Equation of a line parallel to a given line -

$Ax+By+\lambda =0$ is line parallel to $Ax+By+C =0$ .

- wherein

Where $\lambda$ is some constant other than $C$ .

Equation of a line perpendicular to a given line -

$Bx-Ay+\lambda =0$ is the line perpendicular to $Ax+By+C =0$ .

- wherein

$\lambda$ is some other constant than $C$ .

Perpendicular distance of a point from a line -

$\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}$

- wherein

$\rho$ is the distance from the line $ax+by+c=0$ .

To find equations of $AB\:\: and\:\: CD$

$\because AB\:\: and\:\: CD$ are parallel to $3x-4y=0$

and at a distance of $2$ units from $(1,1)$

$\therefore 3x-4y+k=0\:\:and\:\:\left | \frac{3-4+k}{5} \right |=2$

$k-1=\pm 10\:\:\:\:\Rightarrow k=11,-9$

$\therefore$ equations of two sides of the square which are parallel to $3x-4y=0$ are

$3x-4y+11=0\:\:and\:\:3x-4y-9=0$

Now the remaining two sides will be perpendicular to

$3x-4y=0$ and at a distance of $2$ unit from $(1,1)$

$\therefore 4x+3y+k=0$

and $\left | \frac{4+3+k}{5} \right |=2\Rightarrow \:\:\:k+7=\pm 10\:\:\:\:\Rightarrow k=3,-17$

$\therefore$ remaining two sides are

$4x+3y+3=0$ and $4x+3y-17=0$

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