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$\text{ABCD}$ is a rhombus with the diagonals $\text{AC}$ and $\text{BD}$ intersecting at the origin on the $x\text{-}y$ 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 these
in Quantitative Aptitude edited by
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Let’s draw the rhombus first for better understanding.

DIAGRAM

$\text{AD}: x+y = 1 \longrightarrow (1)$

  • put, $x=0 \Rightarrow y=1$
  • put, $y=0 \Rightarrow x=1$

Now, equation of $\text{BC}: y-0 = \frac{0-(-1)}{-1-0}(x+1)$

$\Rightarrow y= -(x+1)$

$\Rightarrow \boxed{x+y=-1}$

(OR)

$y-(-1) = \frac{0-(-1)}{-1-0}(x-0)$

$\Rightarrow y+1 = -x$

$\Rightarrow \boxed{x+y=-1}$

Correct Answer $: \text{A}$

$\text{PS}$

  • Given two points $(x_{1},y_{1})$ and $(x_{2},y_{2}),$ the line passing through these two point is:

$$\boxed{y-y_{1} = \frac{y_{2} – y_{1}}{x_{2} – x_{1}}(x- x_{1})}$$

$$\text{(OR)}$$

$$\boxed{y-y_{2} = \frac{y_{2} – y_{1}}{x_{2} – x_{1}}(x- x_{2})}$$

  • Properties of rhombus: A rhombus is a quadrilateral whose sides are all equal.
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