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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

 

 

Posted by

avinash.dongre

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