If a vertex of a square is at (1, –1) and one of its side lie along the line 3x – 4y – 17 = 0 then find the area of the square.

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Let ABCD be the square.Let the coordinates of vertex A be A1,-1.Now, it is given that one of its side  lies along 3x-4y-17=0.Clearly 1,-1 does not satisfy 3x-4y-17=0, so we can say thatAB and AD are not along 3x-4y-17=0.So, either AD or AB is along 3x-4y-17=0.We know that each angle of a square is right angle.Side of square = Length of perpendicular from 1,-1 to the line 3x-4y-17=0     side = ax1+by1+ca2+b2side of square=3.1-4.-1-1732+42=3+4-175=105=2Hence area of square = a2=2×2=4 sq units


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