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The Straight Lines — MATHS STD 11 Science — Question

Gujarat 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.
✓

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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