NCERT Class X Chapter 7: Coordinate Geometry Example 9

NCERT Class X Chapter 7: Coordinate Geometry Example 9

Question:

Find the ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4). Also find the point of intersection.

Given:

Points \( A(5, -6) \) and \( B(-1, -4) \).

To Find:

  • The ratio in which the y-axis divides the line segment AB.
  • The coordinates of the point of intersection.

Formula:

Section formula: The coordinates of the point dividing the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m:n \) are:

$$ \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) $$

Solution:

Step 1: Let the y-axis divide the line segment AB at point P in the ratio \( k:1 \).

Step 2: The coordinates of P using the section formula are:

$$ P = \left( \frac{k \cdot (-1) + 1 \cdot 5}{k+1},\ \frac{k \cdot (-4) + 1 \cdot (-6)}{k+1} \right) $$

Step 3: Since P lies on the y-axis, its x-coordinate is 0.

$$ \frac{k \cdot (-1) + 1 \cdot 5}{k+1} = 0 $$

Step 4: Solve for \( k \):

$$ k \cdot (-1) + 5 = 0 \\ -k + 5 = 0 \\ k = 5 $$

Step 5: Therefore, the required ratio is \( 5:1 \).

Step 6: Find the y-coordinate of the point of intersection:

$$ y = \frac{5 \cdot (-4) + 1 \cdot (-6)}{5+1} = \frac{-20 + (-6)}{6} = \frac{-26}{6} = -\frac{13}{3} $$

Step 7: Thus, the point of intersection is \( (0, -\frac{13}{3}) \).

Result:

The y-axis divides the line segment joining \( (5, -6) \) and \( (-1, -4) \) in the ratio 5:1. The point of intersection is \( (0, -\frac{13}{3}) \).

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