The Straight Lines — MATHS STD 11 Science — Question
Rajasthan 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.
✓
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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