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$25$ ચોરસ એકમ ક્ષેત્રફળવાળા એક ચતુષ્કોણની બે બાજુઓનું સમીકરણ $3x - 4y = 0$ અને $4x + 3y = 0$ છે. ચતુષ્કોણની બાકીની બે બાજુઓનું સમીકરણ :
$3x - 4y \pm 25 = 0, 4x + 3y \pm 25 = 0$
$3x - 4y \pm 5 = 0, 4x + 3y \pm 5 = 0$
$3x - 4y \pm 5 = 0, 4x + 3y \pm 25 = 0$
એકપણ નહિ
Solution
Area of square $=25$
The equation of $2$ sides are $3 x-4 y=0$ and $4 x+3 y=0$
$\therefore a^2=25$
$\Rightarrow a =5$ units.
Since it is a square, opposite sides are parallel. Parallel lines have same equation but for constant.
Hence the other two sides of the square are:
$\Rightarrow 3 x -4 y + m =0$ and $4 x +3 y + n =0$
Distance between $3 x -4 y =0 \quad 3 x -4 y + m =0$ is $5$ unit
$\Rightarrow m =\pm 25$
$\therefore a x+b y+c_1=05 a x+b y+c_2=0$
$\Rightarrow \frac{c_1-c_2}{\sqrt{a_1^2+b^2}}$
$\therefore 3 x-4 y=25=0$ and $4 x+3 y \pm=0$
Hence, the answer is $3 x -4 y \pm 25=0$ and $4 x +3 y =25=0$.
