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A(1, 4), B(3, 2) and C(7, 5) are vertices of a triangle ABC. Find :

(i) the co-ordinates of the centroid of triangle ABC.

(ii) the equation of a line, through the centroid and parallel to AB.

Straight Line Eq

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Answer

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

Substituting values we get,

Centroid =(1+3+73,4+2+53)=(113,113).\text{Centroid } = \Big(\dfrac{1 + 3 + 7}{3}, \dfrac{4 + 2 + 5}{3}\Big) \\[1em] = \Big(\dfrac{11}{3}, \dfrac{11}{3}\Big).

Hence, centroid of triangle = (113,113).\Big(\dfrac{11}{3}, \dfrac{11}{3}\Big).

(ii) Slope of AB=2−43−1=−22=−1.\text{Slope of AB} = \dfrac{2 - 4}{3 - 1} \\[1em] = \dfrac{-2}{2} = -1.

Slope of line parallel to AB will also be equal to -1.

By point-slope form,

Equation of a line, through the centroid and parallel to AB,

⇒ y - y1 = m(x - x1)

⇒y−113=−1(x−113)⇒3y−113=−1×3x−113⇒3y−11=−1(3x−11)⇒3y−11=−3x+11⇒3y+3x=11+11⇒3x+3y=22.\Rightarrow y - \dfrac{11}{3} = -1\Big(x - \dfrac{11}{3}\Big) \\[1em] \Rightarrow \dfrac{3y - 11}{3} = -1 \times \dfrac{3x - 11}{3} \\[1em] \Rightarrow 3y - 11 = -1(3x - 11) \\[1em] \Rightarrow 3y - 11 = -3x + 11 \\[1em] \Rightarrow 3y + 3x = 11 + 11 \\[1em] \Rightarrow 3x + 3y = 22.

Hence, the equation of a line, through the centroid and parallel to AB is 3x + 3y = 22.

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