25 ચોરસ એકમ ક્ષેત્રફળવાળા એક ચતુષ્કોણની બે બા | vedclass

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Question

Area of square $=25$

The equation of $2$ sides are $3 x-4 y=0$ and $4 x+3 y=0$

$	herefore a^2=25$

$\Rightarrow a =5$ units.

Since it is a

Vedclass · Accepted Answer

$3x - 4y \pm 25 = 0, 4x + 3y \pm 25 = 0$

Vedclass · Answer

Area of square $=25$

The equation of $2$ sides are $3 x-4 y=0$ and $4 x+3 y=0$

$	herefore 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

$\Rightarrow m =\pm 25$

$	herefore a x+b y+c_1=05 a x+b y+c_2=0$

$\Rightarrow \frac{c_1-c_2}{\sqrt{a_1^2+b^2}}$

$	herefore 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 $4 x +3 y =25=0$.

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