1. Suppose the two sides of a square are along the straight lines 6x - 8y = 15 and 4y - 3x = 2. Then the area of the square is
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By: anil on 05 May 2019 02.39 am
The two lines are 4y - 3x = 2 ...(1) and 6x - 8y= 15 4y - 3x= -7.5 ...(2) As we can see $$frac{a_{1}}{a_{2}}$$ = $$frac{b_{1}}{b_{2}}$$ $$ eq$$ $$frac{c_{1}}{c_{2}}$$; these two lines are parallel to each other.Hence the distance between these two parallel lines will be the side of the square i.e. d = $$frac{|c_{1} - c_{2}|}{sqrt{a^{2}+b^{2}}}$$ (here a = -3, b = 4 c_{1} = 2 c_{2} = -7.5 ) d = $$frac{2-(-7.5)}{sqrt{(-3)^{2}+(4)^{2}}}$$ = $$frac{9.5}{5}$$ = 1.9 The distance between the parallel lines will be equal to the length of side of the square. $$ herefore$$ Area of square = (1.9)^{2} = 3.61 Sq. units
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eq$$ $$frac{c_{1}}{c_{2}}$$; these two lines are parallel to each other.Hence the distance between these two parallel lines will be the side of the square i.e. d = $$frac{|c_{1} - c_{2}|}{sqrt{a^{2}+b^{2}}}$$ (here a = -3, b = 4 c_{1} = 2 c_{2} = -7.5 ) d = $$frac{2-(-7.5)}{sqrt{(-3)^{2}+(4)^{2}}}$$ = $$frac{9.5}{5}$$ = 1.9 The distance between the parallel lines will be equal to the length of side of the square. $$ herefore$$ Area of square = (1.9)^{2} = 3.61 Sq. units