A line passing through the point P (1, 2) meets the line `x+y=7` at the distance of 3 units from P. Then the slope of this line satisfies the equation

Options

A. `7x^(2)-18x+7=0`
B. `16x^(2)-39x+16=0`
C. `7x^(2)-6x-7=0`
D. `8x^(2)-9x+1=0`

Answer

To solve the problem, we need to find the slope of a line passing through the point P(1, 2) that meets the line \(x + y = 7\) at a distance of 3 units from P.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Point \(P(1, 2)\)
- Line equation: \(x + y = 7\)
- Distance from point P to the line: 3 units

2. **Set Up the Equation of the Line:**
- The line passing through point P with slope \(m\) can be expressed as:
\[ y - 2 = m(x - 1) \]
- Rearranging gives:
\[ mx - y + (2 - m) = 0 \]

3. **Distance from Point to Line Formula:**
- The distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \(A = m\), \(B = -1\), \(C = 2 - m\), and the point is \(P(1, 2)\).

4. **Substitute into the Distance Formula:**
- The distance from point P to the line is 3:
\[ 3 = \frac{|m(1) - 1(2) + (2 - m)|}{\sqrt{m^2 + (-1)^2}} \]
- Simplifying the numerator:
\[ |m - 2 + 2 - m| = |0| = 0 \]
- Therefore, the equation becomes:
\[ 3 = \frac{|0|}{\sqrt{m^2 + 1}} \quad \text{(This is incorrect; let's fix it)} \]

5. **Correcting the Distance Calculation:**
- The correct distance should be calculated using the coordinates of the point we find 3 units away from P. The coordinates of the point at a distance of 3 units from P can be expressed as:
\[ (1 \pm 3\cos\theta, 2 \pm 3\sin\theta) \]
- We need to ensure this point lies on the line \(x + y = 7\):
\[ (1 + 3\cos\theta) + (2 + 3\sin\theta) = 7 \quad \text{(taking the positive case)} \]
- This simplifies to:
\[ 3 + 3\cos\theta + 3\sin\theta = 7 \]
- Rearranging gives:
\[ \cos\theta + \sin\theta = \frac{4}{3} \]

6. **Square Both Sides:**
- Squaring both sides:
\[ (\cos\theta + \sin\theta)^2 = \left(\frac{4}{3}\right)^2 \]
- Expanding gives:
\[ \cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta = \frac{16}{9} \]
- Since \(\cos^2\theta + \sin^2\theta = 1\):
\[ 1 + 2\cos\theta\sin\theta = \frac{16}{9} \]
- Rearranging gives:
\[ 2\cos\theta\sin\theta = \frac{16}{9} - 1 = \frac{7}{9} \]

7. **Using the Double Angle Formula:**
- Recall that \(2\cos\theta\sin\theta = \sin(2\theta)\):
\[ \sin(2\theta) = \frac{7}{9} \]

8. **Relate to Slope:**
- The slope \(m = \tan\theta\). We can use the identity:
\[ \sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} \]
- Setting this equal to \(\frac{7}{9}\):
\[ \frac{2m}{1 + m^2} = \frac{7}{9} \]

9. **Cross-Multiply and Rearrange:**
- Cross-multiplying gives:
\[ 18m = 7 + 7m^2 \]
- Rearranging leads to:
\[ 7m^2 - 18m + 7 = 0 \]

10. **Final Equation:**
- The slope \(m\) satisfies the quadratic equation:
\[ 7m^2 - 18m + 7 = 0 \]