For an equilateral triangle ΔABC with vertices, A(1, 2), B(2, 3), its

1.	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) \(\bigg(\frac{3+\sqrt3}{6},\frac{5-\sqrt3}{6}\bigg)\)
Answer» (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 \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\).

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

Answer»

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