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The Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsThe Straight Lines3 Marks
Question
Prove that the lines $\sqrt{3}\text{x}+\text{y}=0,\sqrt{3}\text{y}+\text{x}=0,\sqrt{3}\text{x}+\text{y}=1$ and $\sqrt{3}\text{y}+\text{x}=1$ form a rhombus.
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Answer

Let ABCD be a quadrilateral with sides AB, BC, CD, BA as $\sqrt3\text{x+y} = 0, \sqrt3\text{y}+\text{x}=0,\sqrt3\text{x+y}=1$ and $\sqrt{3}\text{y}+\text{x}=1$ respectively.
The slope of $\text{AB}=-\sqrt3 \ ...(1)$
The slope of $\text{BC}=\frac{-1}{\sqrt3} \ ...(2)$
The slope of $\text{CD}=-\sqrt3 \ ...(3)$
The slope of $\text{DA}=\frac{-1}{\sqrt3} \ ...(4)$
From (1), (2), (3) and (4) we observe the slope of opposite sides of quadrilateral are equal.
$\therefore$ Opposite sides are parrallel.
$\therefore$ ABCD is a parallelogram.
We observe that distance between (AD and BC) and (DC and AB) is equal = 1 unit
$\therefore$ Sides AD = AB = BC = DC
$\therefore$ The given figure ABCD is a rhombus
Hence, proved

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