TS EAMCET · Maths · Straight Lines
An equilateral triangle is constructed between the lines \(\sqrt{3} x+y-6=0\) and \(\sqrt{3} x+y+9=0\) with base on one line and vertex on the other. The area (in sq. units) of the triangle so formed is
- A \(\frac{175}{6 \sqrt{3}}\)
- B \(\frac{225}{2 \sqrt{3}}\)
- C \(\frac{225}{4 \sqrt{3}}\)
- D \(\frac{245}{4 \sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{225}{4 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Distance between parallel lines \( =\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2}}=\frac{|9+6|}{\sqrt{3+1}}=\frac{15}{2} \) Also for equilateral triangle…
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