Q) Find the coordinates of the centroid P of the Δ 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.

ICSE Specimen Question Paper (SQP)2025

Ans:

(a) Coordinates of the Centroid:

Given that two vertices of a triangle are A(-1, 3), B(3, -1) and C(0, 0).

Let the centroid coordinates be (X, Y)

We know that if a triangle has 3 coordinates as (x1, y1), (x2, y2), (x3, y3);

then the centroid is given by: (\frac{(x_1 + x_2 + x_3)}{3},\frac{(y_1 + y_2 + y_3)}{3})

∴ for coordinates of Centroid = (\frac{((-1) + 3 + 0)}{3}, \frac{(3 + (-1) + 0)}{3})

= (\frac{2}{3}, \frac{2}{3})

Therefore, the coordinates of the centroid are: (\frac{2}{3}, \frac{2}{3})

(b) Equation of the line parallel to AB:

Step 1: Let’s first calculate gradient or slope of the line AB

We know that the slope of a line passing through points (x1, y1) and (x2, y2) is given by:

m = \frac{y_2 - y_1}{x_2 - x_1}

∴ slope of line AB, mAB passing through points  A(-1, 3), B(3, -1) will be:

mAB = \frac{(- 1) -  (3)}{(3) - (- 1)}

= \frac{- 4}{4}

= -1

Step 2: Now the line parallel to AB will have same slop i.e. its m = – 1

(note: if slope of 2nd line is not same as 1st line, it will intersect the 1st line; which should not be the case of parallel lines)

Hence, the equation of the line parallel to AB and passing through point (x1, y1) is given by:

y – y1 = m (x – x1)

Step 3: Since it is given that this line is passing through the centroid and centroid coordinates are (\frac{2}{3}, \frac{2}{3})

∴ y – \frac{2}{3} = (- 1) ( x – \frac{2}{3})

∴ y – \frac{2}{3} = –  x + \frac{2}{3}

∴ y + x = \frac{4}{3}

∴ 3 x + 3 y = 4

Therefore, the equation of the line is 3 x + 3 y = 4. 

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