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

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Problem
We are given a point \( P(1, 2) \) and a line \( x + y = 7 \). We need to find the slope of a line that passes through point \( P \) and meets the line \( x + y = 7 \) at a distance of 3 units from point \( P \).

### Step 2: Use the Distance Formula
The distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x + y = 7 \), we can rewrite it in the form \( Ax + By + C = 0 \) as:
\[ x + y - 7 = 0 \quad \Rightarrow \quad A = 1, B = 1, C = -7 \]
Substituting \( P(1, 2) \) into the distance formula:
\[ d = \frac{|1 \cdot 1 + 1 \cdot 2 - 7|}{\sqrt{1^2 + 1^2}} = \frac{|1 + 2 - 7|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]
This distance is not equal to 3, so we need to find a point on the line \( x + y = 7 \) that is 3 units away from \( P(1, 2) \).

### Step 3: Parametric Representation of the Line
Let the slope of the line through \( P \) be \( m \). The equation of the line can be expressed as:
\[ y - 2 = m(x - 1) \quad \Rightarrow \quad y = mx - m + 2 \]
This line intersects \( x + y = 7 \). Substituting \( y \) into the line equation:
\[ x + (mx - m + 2) = 7 \]
\[ (1 + m)x - m + 2 = 7 \quad \Rightarrow \quad (1 + m)x = 5 + m \quad \Rightarrow \quad x = \frac{5 + m}{1 + m} \]
Substituting \( x \) back to find \( y \):
\[ y = 7 - x = 7 - \frac{5 + m}{1 + m} = \frac{7(1 + m) - (5 + m)}{1 + m} = \frac{7 + 7m - 5 - m}{1 + m} = \frac{2 + 6m}{1 + m} \]

### Step 4: Calculate the Distance
Now we have the intersection point \( \left( \frac{5 + m}{1 + m}, \frac{2 + 6m}{1 + m} \right) \). The distance from \( P(1, 2) \) to this point should equal 3:
\[ \sqrt{\left( \frac{5 + m}{1 + m} - 1 \right)^2 + \left( \frac{2 + 6m}{1 + m} - 2 \right)^2} = 3 \]
Squaring both sides:
\[ \left( \frac{5 + m - (1 + m)}{1 + m} \right)^2 + \left( \frac{2 + 6m - 2(1 + m)}{1 + m} \right)^2 = 9 \]
This simplifies to:
\[ \left( \frac{4}{1 + m} \right)^2 + \left( \frac{4m}{1 + m} \right)^2 = 9 \]
\[ \frac{16 + 16m^2}{(1 + m)^2} = 9 \]
Cross-multiplying gives:
\[ 16 + 16m^2 = 9(1 + 2m + m^2) \quad \Rightarrow \quad 16 + 16m^2 = 9 + 18m + 9m^2 \]
Rearranging gives:
\[ 7m^2 - 18m + 7 = 0 \]

### Step 5: Final Equation
Thus, the slope \( m \) satisfies the equation:
\[ 7m^2 - 18m + 7 = 0 \]