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

Let the three points dividing AB into four equal parts be P₁, P₂, P₃.

P₁ divides AB in ratio 1:3:

$$P_1 = \left(\frac{1(2)+3(-2)}{1+3},\ \frac{1(8)+3(2)}{1+3}\right) = \left(\frac{2-6}{4},\ \frac{8+6}{4}\right) = \left(-1,\ \frac{7}{2}\right)$$

P₂ divides AB in ratio 2:2 (midpoint):

$$P_2 = \left(\frac{-2+2}{2},\ \frac{2+8}{2}\right) = (0,\ 5)$$

P₃ divides AB in ratio 3:1:

$$P_3 = \left(\frac{3(2)+1(-2)}{3+1},\ \frac{3(8)+1(2)}{3+1}\right) = \left(\frac{4}{4},\ \frac{26}{4}\right) = \left(1,\ \frac{13}{2}\right)$$

The three points are $\left(-1,\ \dfrac{7}{2}\right)$, $(0,\ 5)$, and $\left(1,\ \dfrac{13}{2}\right)$.

Source: Chapter 7, Section 7.3 (Section Formula)

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• To divide a segment into 4 equal parts, you need 3 points dividing it in ratios 1:3, 2:2, and 3:1.
• Apply the section formula: $\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\ \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$.
• P₂ is simply the midpoint — a shortcut saves calculation time.
• Show all three calculations clearly; examiners award marks for each correct point (1 mark each).