Find the ratio in which the line 3x + 4y = 7 divides the line segment joining the points (1, 2) and (– 2, 1).
(a)3:8
(b)4:9
(c)4:8
(d)9:4

Answer

To find the ratio in which the line \(3x + 4y = 7\) divides the line segment joining the points \(A(1, 2)\) and \(B(-2, 1)\), we can use the section formula.

### Step-by-step Solution:

1. **Identify the Points**:
- Let \(A(1, 2)\) and \(B(-2, 1)\).
- We need to find the point \(P(x, y)\) that divides the segment \(AB\) in the ratio \(k:1\).

2. **Apply the Section Formula**:
- The coordinates of point \(P\) that divides the line segment \(AB\) in the ratio \(k:1\) are given by:
\[ x = \frac{k \cdot x_2 + 1 \cdot x_1}{k + 1} = \frac{k \cdot (-2) + 1 \cdot 1}{k + 1} = \frac{-2k + 1}{k + 1} \]
\[ y = \frac{k \cdot y_2 + 1 \cdot y_1}{k + 1} = \frac{k \cdot 1 + 1 \cdot 2}{k + 1} = \frac{k + 2}{k + 1} \]

3. **Substitute into the Line Equation**:
- The point \(P(x, y)\) must satisfy the line equation \(3x + 4y = 7\).
- Substitute \(x\) and \(y\) into the equation:
\[ 3\left(\frac{-2k + 1}{k + 1}\right) + 4\left(\frac{k + 2}{k + 1}\right) = 7 \]
- Simplifying this gives:
\[ \frac{3(-2k + 1) + 4(k + 2)}{k + 1} = 7 \]
\[ \frac{-6k + 3 + 4k + 8}{k + 1} = 7 \]
\[ \frac{-2k + 11}{k + 1} = 7 \]

4. **Cross-Multiply and Solve for \(k\)**:
- Cross-multiplying gives:
\[ -2k + 11 = 7(k + 1) \]
\[ -2k + 11 = 7k + 7 \]
\[ 11 - 7 = 7k + 2k \]
\[ 4 = 9k \]
\[ k = \frac{4}{9} \]

5. **Determine the Ratio**:
- The ratio in which the line divides the segment \(AB\) is \(k:1 = \frac{4}{9}:1\).
- This can be expressed as \(4:9\).

### Final Answer:
The ratio in which the line \(3x + 4y = 7\) divides the line segment joining the points \( (1, 2) \) and \( (-2, 1) \) is \(4:9\).