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Show that the four lines ax\pm by\pm c=0\, form a rhombus whose area is {\frac  {2c^{2}}{ab}}\,.

Given lines are ax+by+c=0,ax+by-c=0,ax-by+c=0,ax-by-c=0\,.Let these lines be 1,2,3,4.

Solving 1 with 3 and 4,we get the vertices A(\left({\frac  {-c}{a}},0\right),B(\left(0,{\frac  {-c}{b}}\right)\,

Solving 2 with 3 and 4, we get D(\left(0,{\frac  {c}{b}}\right),C(\left({\frac  {c}{a}},0\right)\,

Therefore Diagonal AC={\frac  {2c}{a}},BD={\frac  {2c}{b}}\,.Clearly AC lies along x-axis and BD lies along y-axis.They are perpendicular and hence rhombus.

Area={\frac  {1}{2}}|(AC)(BD)|={\frac  {1}{2}}|[{\frac  {2c}{a}}][{\frac  {2c}{b}}|={\frac  {2c^{2}}{|ab|}}\,

Main Page:Geometry:Straight Lines-I