The opposite vertices of a square are 2 0 and 5 1 find opposite vertices
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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) =√[(-2-3)2 + (3-4)2] = (-5)2 + (-1)2 ⇒ 26 Distance between (3, 4) and (x, y) = √[(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) = √[(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 ⇒ x2 - 6x + (6 - 5x)2 - 8(6 - 5x) + 25 = 26 ⇒ x2 - 6x + 36 + 25x2 - 60x - 48 + 40x + 25 = 26 ⇒ 26x2 - 26x - 13 = 0 ⇒ 2x2 - 2x - 1 = 0 Solving the quadratic equation using the quadratic formula, [-b ± √(b2 - 4ac)]/2a. x = [2 ± √(4+8)] / 4 x = [2 ± √(12)] / 4 x = [2 ± 2√(3)] / 4 x = [1 ± √(3)] / 2 y = (6 - 5x) = 6 - 5 [1 ± √(3)] / 2 = [12 - 5 ± 5√(3)] / 2 = [7 ± 5√(3)] / 2 Hence, the coordinates of the third vertex of the equilateral triangle are ([1 ± √(3)] / 2, [7 ± 5√(3)] / 2).