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Question
If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle.
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Solution
Equation of the base AB of a ΔABC is x + y = 2
In ΔABD,
sin 60° = `"AD"/"AB"`
⇒ `sqrt(3)/2 = "AD"/"AB"`
⇒ AD = `sqrt(3)/2 "AB"`
Length of perpendicular from A(2, – 1) to the line x + y = 2 is
AD = `|(1 xx 2 + 1 xx -1 - 2)/sqrt((1)^2 + (1)^2)|`
⇒ `sqrt(3)/2 "AB" = |(2 - 1 - 2)/sqrt(2)| = |(-1)/sqrt(2)|`
⇒ `sqrt(3)/2 "AB" = 1/sqrt(2)`
⇒ AB = `sqrt(2)/sqrt(2)`
Hence, the required length of side = `sqrt(2/3)`.
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