Question

Determine the ratio in which the line \(2x + y = 6\) divides the line segment joining the points (1, 3) and (2, 5).

Hint:

A positive value of \(k\) implies internal division, while a negative value of \(k\) would represent external division.
Since we got \(k = \frac{1}{3} \gt 0\), it is confirmed as internal division.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
We are given two points \(A(1, 3)\) and \(B(2, 5)\).
A straight line \(2x + y = 6\) intersects the line segment joining these points.
We need to find the ratio in which this line divides the segment \(AB\).

Step 2: Key Formula or Approach:
1. Let the line divide \(AB\) in the ratio \(k : 1\) at point \(P\).
2. Apply the Section Formula to find coordinates of \(P\):
\[ P = \left( \frac{k x_2 + x_1}{k + 1}, \frac{k y_2 + y_1}{k + 1} \right) \]
3. Since point \(P\) lies on the line \(2x + y = 6\), its coordinates must satisfy the line equation. We will substitute the coordinates of \(P\) into the equation and solve for \(k\).

Step 3: Detailed Explanation:

• Let the division ratio be \(k : 1\).
- Given points are \(A(1, 3)\) and \(B(2, 5)\).

• Apply the section formula to find coordinates of intersection point \(P\):
\[ x_P = \frac{k(2) + 1}{k + 1} = \frac{2k + 1}{k + 1} \] \[ y_P = \frac{k(5) + 3}{k + 1} = \frac{5k + 3}{k + 1} \]

• Substitute \(x_P\) and \(y_P\) into the line equation \(2x + y = 6\):
\[ 2\left(\frac{2k + 1}{k + 1}\right) + \frac{5k + 3}{k + 1} = 6 \]

• Multiply both sides by \((k + 1)\) to clear the denominators:
\[ 2(2k + 1) + 5k + 3 = 6(k + 1) \] \[ 4k + 2 + 5k + 3 = 6k + 6 \] \[ 9k + 5 = 6k + 6 \]

• Solve for \(k\):
\[ 9k - 6k = 6 - 5 \] \[ 3k = 1 \implies k = \frac{1}{3} \] - This means the ratio is \(\frac{1}{3} : 1\), which simplifies to 1 : 3.


Step 4: Final Answer:
The line divides the segment joining the points in the ratio 1 : 3 internally.