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Maharashtra State BoardSSC (English Medium) 10th Standard

Show that A(1, 2), (1, 6), C(1 + 2sqrt(3), 4) are vertices of an equilateral triangle.

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Question

Show that A(1, 2), (1, 6), C(1 + 2`sqrt(3)`, 4) are vertices of an equilateral triangle.

Sum
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Solution

Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

By distance formula,

AB = `sqrt((1- 1)^2 + (6 - 2)^2`

= `sqrt(0^2 + 4^2)`

= `sqrt(4^2)`

= 4   ...(i)

BC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 - 6)^2`

= `sqrt((2sqrt(3))^2 + (-2)^2`

= `sqrt(12 + 4)`

= `sqrt(16)`

= 4   ...(ii)

AC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 -2)^2`

= `sqrt((2sqrt(3))^2 + 2^2`

= `sqrt(12 + 4)`

= `sqrt(16)`

= 4   ...(iii)

∴ AB = BC = AC   ...[From (i), (ii) and (iii)]

∴ ∆ABC is an equilateral triangle.

∴ Points A, B and C are the vertices of an equilateral triangle.

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Chapter 5: Co-ordinate Geometry - Q.4

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