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A(-1, 3), B(4, 2), C(3, -2) are the vertices of a triangle.

(i) Find the coordinates of the centroid G of the triangle.

(ii) Find the equation of the line through G and parallel to AC.

Straight Line Eq

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Answer

(i) Centroid of the triangle is given by,

G(x,y)=(x1+x2+x33,y1+y2+y33)=(−1+4+33,3+2−23)=(63,33)=(2,1).G(x, y) = \Big(\dfrac{x1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big) \\[1em] = \Big(\dfrac{-1 + 4 + 3}{3}, \dfrac{3 + 2 - 2}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] = (2, 1).

Hence, the coordinates of the centroid G of the triangle is (2, 1).

(ii) Slope of AC = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

=−2−33−(−1)=−54.= \dfrac{-2 - 3}{3 - (-1)} \\[1em] = -\dfrac{5}{4}.

So, the slope of the line parallel to AC is also −54.-\dfrac{5}{4}. and it passes through (2, 1). Hence, its equation can be given by point-slope form i.e.,

⇒y−y1=m(x−x1)⇒y−1=−54(x−2)⇒4(y−1)=−5(x−2)⇒4y−4=−5x+10⇒4y+5x=14⇒5x+4y−14=0.\Rightarrow y - y1 = m(x - x1) \\[1em] \Rightarrow y - 1 = -\dfrac{5}{4}(x - 2) \\[1em] \Rightarrow 4(y - 1) = -5(x - 2) \\[1em] \Rightarrow 4y - 4 = -5x + 10 \\[1em] \Rightarrow 4y + 5x = 14 \\[1em] \Rightarrow 5x + 4y - 14 = 0.

Hence, the equation of the required line is 5x + 4y - 14 = 0.

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