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in Coordinate Geometry by (23.5k points)
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For an equilateral triangle ΔABC with vertices, A(1, 2), B(2, 3), its incentre is \(\bigg(\frac{9+\sqrt3}{6},\frac{15-\sqrt3}{6}\bigg)\). The coordinates of vertex C are:

(a) \(\bigg(\frac{3-\sqrt3}{2},\frac{5+\sqrt3}{2}\bigg)\)

(b) \(\bigg(\frac{3-\sqrt3}{6},\frac{5+\sqrt3}{6}\bigg)\)

(c) \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\)

(d) \(\bigg(\frac{3+\sqrt3}{6},\frac{5-\sqrt3}{6}\bigg)\)

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(c) \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\)

Let the co-ordinates of vertex C are (x, y). 

Since the incentre and centroid of an equilateral triangle coincide, co-ordinates of centroid of

ΔABC = \(\bigg(\frac{9+\sqrt3}{6},\frac{15-\sqrt3}{6}\bigg)\)

∴ \(\frac{x+1+2}{3}\) = \(\frac{9+\sqrt3}{6}\) and \(\frac{2+3+y}{3}\) = \(\frac{15-\sqrt3}{6}\)

⇒ 2(x + 3) = 9 + √3 and 2(5 + y) = 15 – √3 

⇒ 2x = 3 + √3 and 10 + 2y = 15 – √3

⇒ x = \(\frac{3+\sqrt3}{2}\) and 2y = 5 - √5 ⇒ y = \(\frac{5-\sqrt3}{2}\)

∴ Required co-ordinates are \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\).

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