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$If the quadrilateral formed by the lines a x + b y + c = 0, a^{'} x + b^{'} y + c = 0, a x + b y + c^{'} = 0, a^{'} x + b^{'} y + c^{'} = 0 has$
$perpendicular diagonals, then$

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b2+c2=b′2+c′2

c2+a2=c′2+a′2

a2+b2=a′2+b′2

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detailed solution

Correct option is C

Since the diagonals are perpendicular, the given quadrilateral is a rhombus. So, the distances between two pairs of parallel sides are equal. Hence,c′−ca2+b2=∣c′−ca′2+b′2 or a2+b2=a12+b12

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If the quadrilateral formed by the lines ax+by+c=0 , a′x+b′y+c=0,ax+by+c′=0,a′x+b′y+c′=0 has perpendicular diagonals, then

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$If the quadrilateral formed by the lines a x + b y + c = 0, a^{'} x + b^{'} y + c = 0, a x + b y + c^{'} = 0, a^{'} x + b^{'} y + c^{'} = 0 has$
$perpendicular diagonals, then$