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The Straight Lines — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsThe Straight Lines5 Marks
Question
Prove that the lines $\text{y}=\sqrt{3}\text{x}+1,\text{y}=4$ and $\text{y}=-\sqrt{3}\text{x}+2$ form an equilateral triangle.
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Answer

The given equations are as follows:
$\text{y}=\sqrt3\text{x}+1\ ...(1)$
$\text{y}=4\ ...(2)$
$\text{y} = -\sqrt{3}\text{x}+2\ ...(3)$
In triangle ABC, let equation (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2)
$\text{x}= \sqrt3,\text{y}=4$
Thus, AB and BC intersect at $\text{B}\big(\sqrt3,4\big)$
Solving (1) and (3)
$\text{x} = \frac{1}{2\sqrt3}, \text{y} = \frac{3}{2}$
Thus, AB and CA intersect at $\text{A}\big(\frac{1}{2\sqrt3},\frac{3}{2}\big).$
Similarly, solving (2) and (3)
$\text{x} = -\frac{2}{\sqrt3}, \text{y}= 4$
Thus, BC and AC intersect at $\text{C}\big(-\frac{2}{\sqrt{3}},4\big).$
Now, we have:
$\text{AB}=\sqrt{\Big(\frac{1}{2\sqrt{3}}-\sqrt{3}\Big)^2+\Big(\frac{3}{2}-4\Big)^2}=\frac{5}{\sqrt3}$
$\text{BC}=\sqrt{\Big(\sqrt{3}+\frac{2}{\sqrt3}\Big)^2+\Big(4-4\Big)^2}=\frac{5}{\sqrt3}$
$\text{AC}=\sqrt{\Big(\frac{1}{2\sqrt3}+\frac{2}{\sqrt3}\Big)^2+\Big(\frac{3}{2}-4\Big)^2}=\frac{5}{\sqrt3}$
Hence, the given lines form an equilateral triangle.

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