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Question
If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
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Solution
Let the vertices be (x, y)
Distance between (x, y) and (4, 3) is = `sqrt((x - 4)^2 + (y - 3)^2)` ...(1)
Distance between (x,y) and (– 4, 3) is = `sqrt((x + 4)^2 + (y - 3)^2)` ...(2)
Distance between (4, 3) and (– 4, 3) is = `sqrt((4 + 4)^2 + (3 - 3)^2) = sqrt(8)^2`= 8
According to the question,
Equation (1) = (2)
(x – 4)2 = (x + 4)2
x2 – 8x + 16 = x2 + 8x + 16
16x = 0
x = 0
Also, equation (1) = 8
(x – 4)2 + (y – 3)2 = 64 ...(3)
Substituting the value of x in (3)
Then (0 – 4)2 + (y – 3)2 = 64
(y – 3)2 = 64 – 16
(y – 3)2 = 48
y – 3 = `(+)4sqrt(3)`
y = `3(+) 4sqrt(3)`
Neglect y = `3(+) 4sqrt(3)` as if y = `3(+) 4sqrt(3)` then origin cannot interior of triangle
Therefore, the third vertex = `(0, 3 - 4sqrt(3))`
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