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Question

Show that the four lines ax±by±c=0 enclose a rhombus whose area is 2c2ab.

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Solution

The four sides of the rhombus are
a+xby+c=0
ax+byc=0
axby+c=0
axbyc=0
Solving (1) with (3) and (2) with (4), we get the vertices (c/a,0),(c/a,0).
Solving (2) with (3) and (1) with (4),we get the vertices as (0,c/b)(0,c/b)
Above shown that one diagonal is of length 2c/a and is along xaxis whereas the other is along yaxis and is length 2c/b. Since the diagonals being along the axes, they are perpendicular and hence a rhombus.
The area is 12(d1.d2)=12(2ca.2cb)=2c2ab
Note : You may do it as in Q.25.
Area=p1p2sinθ=∣ ∣ ∣ ∣(c1d1)(c2d2)a1b1a2b2∣ ∣ ∣ ∣
=∣ ∣ ∣ ∣(2c)(2c)abab∣ ∣ ∣ ∣=4c22ab=2c2ab

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