NCERT Class X Chapter 7: Coordinate Geometry Example 8
NCERT Class X Chapter 7: Coordinate Geometry Example 8
Question:
Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, –2) and B(–7, 4).
Given:
Points \( A(2, -2) \) and \( B(-7, 4) \).
To Find:
Coordinates of the points of trisection of the line segment \( AB \).
Formula:
Section formula: If a point \( P \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then the coordinates of \( P \) are:
\[ P = \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) \]
Solution:
Step 1: Let the points of trisection be \( P \) and \( Q \). Point \( P \) divides \( AB \) in the ratio \( 1:2 \), and point \( Q \) divides \( AB \) in the ratio \( 2:1 \).
Step 2: Find the coordinates of \( P \) (dividing \( AB \) in \( 1:2 \)).
\[ \begin{align*} x_P &= \frac{1 \times (-7) + 2 \times 2}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1 \\ y_P &= \frac{1 \times 4 + 2 \times (-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0 \\ \end{align*} \]So, \( P = (-1, 0) \).
Step 3: Find the coordinates of \( Q \) (dividing \( AB \) in \( 2:1 \)).
\[ \begin{align*} x_Q &= \frac{2 \times (-7) + 1 \times 2}{2 + 1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4 \\ y_Q &= \frac{2 \times 4 + 1 \times (-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2 \\ \end{align*} \]So, \( Q = (-4, 2) \).
Result:
The coordinates of the points of trisection are \( (-1, 0) \) and \( (-4, 2) \).
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