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Problem
Problem
Show that the area of the parallelogram formed by the lines
2
x
−
3
y
+
a
=
0
2x - 3y + a = 0
2
x
−
3
y
+
a
=
0
,
3
x
−
2
y
−
a
=
0
3x - 2y - a = 0
3
x
−
2
y
−
a
=
0
,
2
x
−
3
y
+
3
a
=
0
2x - 3y + 3a = 0
2
x
−
3
y
+
3
a
=
0
and
3
x
−
2
y
−
2
a
=
0
3x - 2y - 2a = 0
3
x
−
2
y
−
2
a
=
0
is
2
a
2
5
\frac{2a^2}{5}
5
2
a
2
sq units.
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