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In the adjoining diagram, G is the centroid of △ ABC. A(3, -3), B(2, -6), C(x, y) and G(5, -5). The coordinates of point D are :

  1. (2, -6)

  2. (3, -6)

  3. (6, -6)

  4. (10, -6)

In the adjoining diagram, G is the centroid of △ ABC. A(3, -3), B(2, -6), C(x, y) and G(5, -5). The coordinates of point D are : Maths Competency Focused Practice Questions Class 10 Solutions.

Section Formula

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Answer

By formula,

Centroid of triangle = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

∴(5,−5)=(3+2+x3,(−3)+(−6)+y3)⇒(5,−5)=(x+53,y−93)⇒x+53=5 and y−93=−5⇒x+5=15 and y−9=−15⇒x=15−5=10 and y=−15+9=−6.\therefore (5, -5) = \Big(\dfrac{3 + 2 + x}{3}, \dfrac{(-3) + (-6) + y}{3}\Big) \\[1em] \Rightarrow (5, -5) = \Big(\dfrac{x + 5}{3}, \dfrac{y - 9}{3}\Big) \\[1em] \Rightarrow \dfrac{x + 5}{3} = 5 \text{ and } \dfrac{y - 9}{3} = -5 \\[1em] \Rightarrow x + 5 = 15 \text{ and } y - 9 = -15 \\[1em] \Rightarrow x = 15 - 5 = 10 \text{ and } y = -15 + 9 = -6.

C(x, y) = (10, -6).

Since, centroid is the point of intersection of all the three medians of a triangle.

∴ AD is the median.

∴ D is mid-point of BC.

D=(2+102,(−6)+(−6)2)=(122,−122)=(6,−6).D = \Big(\dfrac{2 + 10}{2}, \dfrac{(-6) + (-6)}{2}\Big) \\[1em] = \Big(\dfrac{12}{2}, \dfrac{-12}{2}\Big) \\[1em] = (6, -6).

Hence, Option 3 is the correct option.

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