Find the coordinates of the points of trisection of the line

Q) Find the coordinates of the points of trisection of the line segment joining the points A(- 1, 4) and B(- 3, – 2)

(Q 30 – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Since, A Trisection means dividing a line segment into three equal parts.

It means, if AB is a line, then P and Q are two points on AB, such that:

AP = PQ = QB

∴ we can say that P divides the line AB in 1 : 2

Similarly, we can say that Q divides the line AB in 2 : 1

Step 2: We know that the coordinates of a point, dividing the line, in ration m:n, is given by

$(\frac{m X_2 + n X_1}{m + n}, \frac{m Y_2 + n Y_1}{m + n})$

From 1st condition, P divides line AB in ration of 1 : 2

Here, coordinates of point A are (- 1, 4) and B (- 3, – 2)

∴ coordinates of a point P, are: $(\frac{1 (-3) + 2 (- 1)}{1 + 2}, \frac{1 (- 2) + 2 (4)}{1 + 2})$

= $(\frac{(-3) + (- 2)}{3}, \frac{(- 2) + (8)}{3})$

= $(\frac{-5}{3}, \frac{6}{3})$

= $(\frac{-5}{3}, 2)$

Step 3: From 2nd condition, Q divides line AB in ration of 2 : 1,

Here, coordinates of point A are (- 1, 4) and B (- 3, – 2)

∴ coordinates of a point Q, are: $(\frac{2 (-3) + 1 (- 1)}{2 + 1}, \frac{2 (- 2) + 1 (4)}{2 + 1})$

= $(\frac{(-6) + (- 1)}{3}, \frac{(- 4) + (4)}{3})$

= $(\frac{- 7}{3}, \frac{0}{3})$

= $(\frac{- 7}{3}, 0)$

Therefore, the coordinates of the points are $(\frac{- 5}{3}, 2)$ and $(\frac{- 7}{3}, 0)$ .

Step 4: (if asked to plot, then) when we plot this line and the trisection points, P and Q, we get this diagram:

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