show that the area of parallelogram formed by the lines 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0 is 2a²/5 square units

Question

Elton · Answer

there is a direct frmula to solve this problem. Area of paralellogram formed by the lines a1x+b1y+c1=0,a2x+b2y+d1=0,a1x+b1y+c2=0 and a2x+b2y+d2=0 is given by:

Pushpam Kumar PK · Answer

Can you solve.... I`m not getting... Please.....,,solve and explain it whole......I wanna pure explanation......

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show that the area of parallelogram formed by the lines 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0 is 2a²/5 square units

show that the area of parallelogram formed by the lines 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0 is 2a²/5 square units

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show that the area of parallelogram formed by the lines 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0 is 2a²/5 square units

show that the area of parallelogram formed by the lines 2x-3y+a=0,3x-2y-a=0,2x-3y+3a=0 and 3x-2y-2a=0 is 2a²/5 square units

2 Answers

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