Two sides of a square with an area of 25 squa | vedclass

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Question

(A) The area of the square is $25$ square units,so the side length $a = \sqrt{25} = 5$ units.The given lines $L_1: 3x - 4y = 0$ and $L_2: 4x + 3y = 0$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The area of the square is $25$ square units,so the side length $a = \sqrt{25} = 5$ units.The given lines $L_1: 3x - 4y = 0$ and $L_2: 4x + 3y = 0$ are perpendicular because their slopes $m_1 = 3/4$ and $m_2 = -4/3$ satisfy $m_1 	imes m_2 = -1$.Since the sides are parallel to the given lines,the other two sides must be of the form $3x - 4y + c_1 = 0$ and $4x + 3y + c_2 = 0$.The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For the first pair of parallel lines ($3x - 4y = 0$ and $3x - 4y + c_1 = 0$):$5 = \frac{|c_1 - 0|}{\sqrt{3^2 + (-4)^2}} = \frac{|c_1|}{5}$ $\Rightarrow |c_1| = 25$ $\Rightarrow c_1 = \pm 25$.For the second pair of parallel lines ($4x + 3y = 0$ and $4x + 3y + c_2 = 0$):$5 = \frac{|c_2 - 0|}{\sqrt{4^2 + 3^2}} = \frac{|c_2|}{5}$ $\Rightarrow |c_2| = 25$ $\Rightarrow c_2 = \pm 25$.Thus,the equations of the other two sides are $3x - 4y \pm 25 = 0$ and $4x + 3y \pm 25 = 0$.

Two sides of a square with an area of $25$ square units are given by the equations $3x - 4y = 0$ and $4x + 3y = 0$. The equations of the other two sides are:

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