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Q.
A square of area $25\, sq$. unit is formed by taking two sides as $3 x +4 y = k _{1}$ and $3 x +4 y = k _{2}$, then find value of $\left |k_{1}-k_{2}\right|$.
Straight Lines
Solution:
Side $d=\left|\frac{k_{1}-k_{2}}{\sqrt{9+16}}\right|$
Area $=25$
$d^{2}=25$
$\left(\frac{ k _{1}- k _{2}}{5}\right)^{2}=25$
$\Rightarrow\left| k _{1}- k _{2}\right|=25$
