NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.2 Question 4

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

Given:

  • Point A: \( (-3, 10) \)
  • Point B: \( (6, -8) \)
  • Point P: \( (-1, 6) \), which divides AB

To Find:

The ratio in which point P divides the line segment AB.

Formula:

Section formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then

\[ x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n} \]

Solution:

Step 1: Let the required ratio be \( m:n \). Using the section formula, the coordinates of the dividing point \( P \) are:

\[ P = \left( \frac{m \cdot 6 + n \cdot (-3)}{m+n},\ \frac{m \cdot (-8) + n \cdot 10}{m+n} \right) \]

Step 2: Since \( P = (-1, 6) \), equate the x-coordinates:

\[ -1 = \frac{6m - 3n}{m+n} \]

Step 3: Cross-multiply and solve for \( m \) and \( n \):

\[ -1(m+n) = 6m - 3n \\ -m - n = 6m - 3n \\ - m - n - 6m + 3n = 0 \\ -7m + 2n = 0 \\ 7m = 2n \\ \frac{m}{n} = \frac{2}{7} \]

Step 4: Now, check with y-coordinates to confirm:

\[ 6 = \frac{-8m + 10n}{m+n} \] \[ 6(m+n) = -8m + 10n \\ 6m + 6n = -8m + 10n \\ 6m + 6n + 8m - 10n = 0 \\ 14m - 4n = 0 \\ 14m = 4n \\ \frac{m}{n} = \frac{4}{14} = \frac{2}{7} \]

Both calculations give the same ratio.

Result:

The line segment joining \( (-3, 10) \) and \( (6, -8) \) is divided by \( (-1, 6) \) in the ratio 2:7.

© Kaliyuga Ekalavya. All rights reserved.

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