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# Find the area of rhombus formed by the lines $x\pm y\pm 1=0$?

$(a)\;2\qquad(b)\;4\qquad(c)\;5\qquad(d)\;7$

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A)
Toolbox:
• Distance between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $\bigg|\large\frac{c_1-c_2}{\sqrt {a^2+b^2}}\bigg|$
The given lines are
$x+y+1=0$--------(1)
$x+y-1=0$--------(2)
$x-y+1=0$--------(3)
$x-y-1=0$--------(4)
Distance between the parallel lines (1) & (2) say $P_1$ is
$P_1=\large\frac{\mid -1-1\mid}{\sqrt{(1)^2+(1)^2}}$
$P_1=\sqrt 2$
Similarly distance between (3) & (4) is also $P_2=\sqrt 2$
Here $P_1=P_2$
$\therefore$ The height of the rhombus$=\sqrt 2$
Solving (1) and (3) we get the point $A(-1,0)$ and
Solving (1) and (4) we get the point $B(0,-1)$
Distance $AB=\sqrt {(1)^2+(-1)^2}=\sqrt 2$
$i.e.,$Base of the rhombus$=\sqrt 2$
Area of the rhombus=$base\times height$$=\sqrt 2\times\sqrt 2=2$
Hence (a) is the correct answer.

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A)
Area will be 2c^2/ab =

2/1=

2..

That simple