If two sides of a square are 4x + 3y - 20 = 0 | vedclass

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(D) The given equations of the lines are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$. Since the coefficients of $x$ and $y$ are the same,the lines are p

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$49$ sq. units

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(D) The given equations of the lines are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$. Since the coefficients of $x$ and $y$ are the same,the lines are parallel. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} \right|$. Here,$a = 4$,$b = 3$,$c_1 = -20$,and $c_2 = 15$. $d = \left| \frac{-20 - 15}{\sqrt{4^2 + 3^2}} \right| = \left| \frac{-35}{\sqrt{16 + 9}} \right| = \left| \frac{-35}{5} \right| = 7$ units. Since the distance between two opposite sides of a square is equal to its side length $s$,we have $s = 7$ units. The area of the square is $s^2 = 7^2 = 49$ sq. units.

If two sides of a square are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$,then the area of the square is

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