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ICSE Class 10 Mathematics Question 14 of 21

Solved 2025 Specimen Paper ICSE Class 10 Mathematics — Question 7

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7
Question

Question 6(i)

Find the coordinates of the centroid P of the △ ABC, whose vertices are A(-1, 3), B(3, -1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.

Answer

By formula,

Centroid of triangle = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

Substituting values we get :

⇒Centroid of △ ABC=(−1+3+03,3+(−1)+03)⇒P=(23,23).\Rightarrow \text{Centroid of △ ABC} = \Big(\dfrac{-1 + 3 + 0}{3}, \dfrac{3 + (-1) + 0}{3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{2}{3}, \dfrac{2}{3}\Big).

By formula,

Slope = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}

Substituting values we get :

Slope of AB=−1−33−(−1)=−44=−1.\text{Slope of AB} = \dfrac{-1 - 3}{3 - (-1)} \\[1em] = \dfrac{-4}{4} \\[1em] = -1.

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y1 = m(x - x1)

Substituting values we get :

Equation of line passing through P and parallel to AB :

⇒y−23=−1(x−23)⇒3y−23=−1×3x−23⇒3y−2=−1(3x−2)⇒3y−2=−3x+2⇒3y+3x=2+2⇒3y+3x=4.\Rightarrow y - \dfrac{2}{3} = -1\Big(x - \dfrac{2}{3}\Big) \\[1em] \Rightarrow \dfrac{3y - 2}{3} = -1 \times \dfrac{3x - 2}{3} \\[1em] \Rightarrow 3y - 2 = -1(3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3y + 3x = 2 + 2 \\[1em] \Rightarrow 3y + 3x = 4.

Hence, required equation is 3x + 3y = 4.