Question

In the following example, can the segment joining the given points form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle.

A(√2, √2), B(−√2, −√2), C(−√6, √6)

Answer 1

By distance formula,

d(A, B) = `sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

d(A, B) = `sqrt((−sqrt2 − sqrt2)^2 + (−sqrt2− sqrt2)^2)`

d(A, B) = `sqrt((−2sqrt2)^2 + (−2sqrt2)^2)`

d(A, B) = `sqrt(8 + 8)`

d(A, B) = `sqrt(16)`

∴ d(A, B) = 4                           ...(i)

d(B, C) = `sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2`

d(B, C) = `sqrt([−sqrt6 − (− sqrt2)]^2 + [sqrt6 − (− sqrt2)]^2)`

d(B, C) = `sqrt((−sqrt6 + sqrt2)^2 + (sqrt6 + sqrt2)^2)`

d(B, C) = `sqrt(6 - 2sqrt12 + 2 + 6 + 2sqrt12 + 2)`

d(B, C) = `sqrt(16)`

∴ d(B, C) = 4                          ...(ii)

d(A, C) = `sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2`

d(A, C) = `sqrt((−sqrt6 − sqrt2)^2 + (sqrt6 − sqrt2)^2)`

d(A, C) = `sqrt(6 + 2sqrt12 + 2 + 6 - 2sqrt12 + 2)`

d(A, C) = `sqrt(16)`

∴ d(A, C) = 4                      ...(iii)

On adding (i) and (ii),

d(A, B) + d(B, C) = 4 + 4 = 8

∴ d(A, B) + d(B, C) + d(A, C)      ...[From (iii)]

∴ Points A, B, C are non collinear points. We can construct a triangle through 3 non collinear points.

∴ The segment joining the given points form a triangle.

Since, AB = BC = AC

∴ ∆ABC is an equilateral triangle.

∴ The segments joining the points A, B and C will form an equilateral triangle.