Geo2.43

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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

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