Mathematics — Standard (041) — CBSE Class 10 board question

Let P and Q be the points of trisection of AB, so AP = PQ = QB.

Point P divides AB in the ratio 1 : 2 internally.

$$P = \left(\frac{1(-3) + 2(-1)}{1+2},\ \frac{1(-2) + 2(4)}{1+2}\right) = \left(\frac{-3-2}{3},\ \frac{-2+8}{3}\right) = \left(\frac{-5}{3},\ 2\right)$$

Point Q divides AB in the ratio 2 : 1 internally.

$$Q = \left(\frac{2(-3) + 1(-1)}{2+1},\ \frac{2(-2) + 1(4)}{2+1}\right) = \left(\frac{-6-1}{3},\ \frac{-4+4}{3}\right) = \left(\frac{-7}{3},\ 0\right)$$

The coordinates of the points of trisection are $\left(\dfrac{-5}{3},\ 2\right)$ and $\left(\dfrac{-7}{3},\ 0\right)$.

Source: Chapter 7, Section 7.3 (Section Formula)

---

• Trisection means dividing into 3 equal parts, giving two points P and Q.
• P divides AB in ratio 1:2 (closer to A), Q divides AB in ratio 2:1 (closer to B).
• Apply the section formula: $x = \dfrac{m_1x_2 + m_2x_1}{m_1+m_2}$, $y = \dfrac{m_1y_2 + m_2y_1}{m_1+m_2}$.
• Always label the two trisection points and show working for both — each point typically carries 1 mark, with 1 mark for method.