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Anuj A

Oct 28, 2014

# Find the third vertex of the equilateral triangle.

An equilateral triangle has vertices (3,4) and (-2,3). Find the third vertex.

Ramesh

Sol: Two vertices of an equilateral triangle are (3, 4) and (-2, 3) Let the third vertex of the triangle be (x, y) Distance between (3, 4) and (-2, 3) =âˆš[(-2-3)2 + (3-4)2] = (-5)2 + (-1)2 â‡’ 26 Distance between (3, 4) and (x, y) = âˆš[(x-3)2 + (y-4)2] = [(x-3)2 + (y-4)2] = [(x-3)2 + (y-4)2] = x2 - 6x + 9 + y2 - 8y + 16  = x2 - 6x + y2 - 8y + 25 ------------------------ (1) Distance between (-2, 3) and (x, y) = âˆš[(x+2)2 + (y-3)2] = [(x+2)2 + (y-3)2] = 26 = x2 +  4x + 4 + y2 - 6y + 9 = x2 + 4x + y2 - 6y + 13 -------------------- (2) Equating the distances we get, x2 - 6x + y2 - 8y + 25 = x2 + 4x + y2 - 6y + 13 10x + 2y - 12 = 0 5x + y - 6 = 0 y = (6 - 5x) Substituting the value of y in equation (1) and equating it to 26. x2 - 6x + y2 - 8y + 25 = 26 â‡’ x2 - 6x + (6 - 5x)2 - 8(6 - 5x) + 25 = 26 â‡’ x2 - 6x + 36 + 25x2 -  60x - 48 + 40x + 25 = 26 â‡’ 26x2 - 26x - 13 = 0 â‡’ 2x2 - 2x - 1 = 0 Solving the quadratic equation using the quadratic formula, [-b Â± âˆš(b2 - 4ac)]/2a. x = [2 Â± âˆš(4+8)] / 4 x = [2 Â± âˆš(12)] / 4 x = [2 Â± 2âˆš(3)] / 4 x = [1 Â± âˆš(3)] / 2 y = (6 - 5x)    = 6 - 5 [1 Â± âˆš(3)] / 2    = [12 - 5 Â± 5âˆš(3)] / 2    = [7 Â± 5âˆš(3)] / 2 Hence, the coordinates of the third vertex of the equilateral triangle are ([1 Â± âˆš(3)] / 2, [7 Â± 5âˆš(3)] / 2).

Syeda

Answer. Let the coordinates of the circumcentre of the triangle be (x, y). Circumcentre of a triangle is equidistant from each of the vertices. Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y) âˆš[(x - 8)2 + (y - 6)2] = âˆš[(x - 8)2 + (y + 2)2] [(x - 8)2 + (y - 6)2] = [(x - 8)2 + (y + 2)2] (y - 6)2 = (y + 2)2 y2 + 36 - 12y = y2 + 4y + 4 36 - 12y = 4y + 4 16y = 32 y = 2 Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y) âˆš[(x - 2)2 + (y + 2)2] = âˆš[(x - 8)2 + (y + 2)2] [(x - 2)2 + (y + 2)2] = [(x - 8)2 + (y + 2)2] (x - 2)2 = (x - 8)2 x2 + 4 - 4x = x2 - 16x + 64 4 - 4x = -16x + 64 12x = 60 x = 5. Hence, the coordiantes of the circumcentre of the triangle are (5, 2). Circumradius = âˆš[(5 - 8)2 + (2 - 6)2] = âˆš(9 + 16) = âˆš25 = 5 units. SME Approved
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