Two sides of a square are along the lines 5x | vedclass

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(A) The given equations of the lines are $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$.Since the coefficients of $x$ and $y$ are proportional,the lines

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$9$ sq. units.

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(A) The given equations of the lines are $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$.Since the coefficients of $x$ and $y$ are proportional,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,$a = 5$,$b = -12$,$c_1 = 39$,and $c_2 = 78$.$d = \left| \frac{39 - 78}{\sqrt{5^2 + (-12)^2}} \right| = \left| \frac{-39}{\sqrt{25 + 144}} \right| = \left| \frac{-39}{\sqrt{169}} \right| = \frac{39}{13} = 3$ units.Since the distance between two parallel sides of a square is equal to its side length,the side of the square is $s = 3$ units.The area of the square is $s^2 = 3^2 = 9$ sq. units.

Two sides of a square are along the lines $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$. The area of the square is:

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