NCERT Class X Chapter 7: Coordinate Geometry Example 7
NCERT Class X Chapter 7: Coordinate Geometry Example 7
Question:
In what ratio does the point (–4, 6) divide the line segment joining the points A(–6, 10) and B(3, –8)?
Given:
A = (–6, 10), B = (3, –8), and P = (–4, 6) divides AB.
To Find:
The ratio in which the 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 for the x-coordinate:
\[ -4 = \frac{m \times 3 + n \times (-6)}{m + n} \]Step 2: Cross multiply and simplify the equation for x-coordinate.
\[ -4(m + n) = 3m - 6n \] \[ -4m - 4n = 3m - 6n \] \[ -4m - 4n - 3m + 6n = 0 \] \[ -7m + 2n = 0 \] \[ 2n = 7m \]Step 3: Express the ratio \( \frac{m}{n} \) from the above equation.
\[ 2n = 7m \implies \frac{m}{n} = \frac{2}{7} \]Step 4: Verify with the y-coordinate using the section formula:
\[ 6 = \frac{m \times (-8) + n \times 10}{m + n} \] \[ 6(m + n) = -8m + 10n \] \[ 6m + 6n = -8m + 10n \] \[ 6m + 6n + 8m - 10n = 0 \] \[ 14m - 4n = 0 \] \[ 14m = 4n \] \[ \frac{m}{n} = \frac{4}{14} = \frac{2}{7} \]Step 5: Therefore, the point (–4, 6) divides AB in the ratio 2:7.
Result:
The point (–4, 6) divides the line segment joining A(–6, 10) and B(3, –8) in the ratio 2:7.
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