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$\text{ABCD}$ is a rhombus with the diagonals $\text{AC}$ and $\text{BD}$ intersecting at the origin on the $x-y$ plane. The equation of the straight line $\text{AD}$ is $x + y = 1.$ What is the equation of $\text{BC}?$

  1. $x + y = –1$
  2. $x – y = –1$
  3. $x + y = 1$
  4. None of the above
in Quantitative Aptitude edited by
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Opposite sides of rhombus are parallel. So their ratios are equal as...

ax+by+c=0

Px+qy+r=0

a/p =b/q then lines are parallel

a/p =b/q =c/r then lines are coincident

Now AD eqn is x+y=1

So it's parallel line BC eqn would be x+y=k

Where k ! =1 otherwise AD and BC would be coincident so

Option 2 and 3 not possible.

And intersecting point of both diagonals is origin so rhombus equally divided by axes

Option 1 can be possible
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