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: Set up the section formula approach.
Let the line $2x + y = 6$ divide the segment joining $A(1, 3)$ and $B(2, 5)$ in the ratio $k:1$.
Step 2: Find the coordinates of the dividing point using the section formula.
\[ x = \frac{k \cdot 2 + 1 \cdot 1}{k + 1} = \frac{2k + 1}{k + 1} \] \[ y = \frac{k \cdot 5 + 1 \cdot 3}{k + 1} = \frac{5k + 3}{k + 1} \]
Step 3: Substitute the point into the line equation.
Since the dividing point lies on $2x + y = 6$: \[ 2 \cdot \frac{2k + 1}{k + 1} + \frac{5k + 3}{k + 1} = 6 \]
Step 4: Simplify and solve for k.
\[ \frac{4k + 2 + 5k + 3}{k + 1} = 6 \implies \frac{9k + 5}{k + 1} = 6 \] \[ 9k + 5 = 6k + 6 \implies 3k = 1 \implies k = \frac{1}{3} \]
Step 5: Write the ratio.
Ratio $= k : 1 = \dfrac{1}{3} : 1 = 1 : 3$.
Step 6: Conclusion.
The line $2x + y = 6$ divides the segment in the ratio $1:3$ internally.
\[ \boxed{1 : 3} \]