MHT CET · Maths · Straight Lines
The base of an equilateral triangle is represented by the equation \(2 x-y-1=0\) and its vertex is \((1,2)\), then the length (in units) of the side of the triangle is
- A \(\sqrt{\frac{20}{13}}\)
- B \(\frac{2}{\sqrt{15}}\)
- C \(\sqrt{\frac{8}{15}}\)
- D \(\sqrt{\frac{15}{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{\sqrt{15}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{AD} & =\left|\frac{2-2-1}{\sqrt{2^2+(-1)^2}}\right| \\ & =\left|\frac{-1}{\sqrt{5}}\right| \\ & =\frac{1}{\sqrt{5}}\end{aligned}\)
\(\begin{aligned} & \text { In } \triangle \mathrm{ABD}, \tan 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{BD}} \\ & \Rightarrow \sqrt{3}=\frac{\frac{1}{\sqrt{5}}}{\mathrm{BD}} \\ & \Rightarrow \mathrm{BD}=\frac{1}{\sqrt{15}} \\ & \therefore \quad \mathrm{BC}=2 \mathrm{BD}=\frac{2}{\sqrt{15}}\end{aligned}\)
\(\begin{aligned} & \text { In } \triangle \mathrm{ABD}, \tan 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{BD}} \\ & \Rightarrow \sqrt{3}=\frac{\frac{1}{\sqrt{5}}}{\mathrm{BD}} \\ & \Rightarrow \mathrm{BD}=\frac{1}{\sqrt{15}} \\ & \therefore \quad \mathrm{BC}=2 \mathrm{BD}=\frac{2}{\sqrt{15}}\end{aligned}\)
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