Let A be the point of intersection of the lines `3x + 2y = 14`, `5x – y = 6` and B be the point of intersection of the lines `4x + 3y = 8`, `6x + y = 5`. The distance of the point `P(5, –2)` from the line AB is

Options

A. `(13/2)`
B. 8
C. `5/2`
D. 6

Answer

To solve the problem, we need to find the points of intersection of the given lines, derive the equation of the line connecting these points, and then calculate the distance from the point P(5, -2) to this line.

### Step 1: Find the intersection point A of the lines \(3x + 2y = 14\) and \(5x - y = 6\).

To find the intersection point A, we will solve the two equations simultaneously.

1. From the first equation, express \(y\) in terms of \(x\):
\[ 2y = 14 - 3x \implies y = \frac{14 - 3x}{2} \]

2. Substitute this expression for \(y\) into the second equation:
\[ 5x - \frac{14 - 3x}{2} = 6 \]

3. Multiply through by 2 to eliminate the fraction:
\[ 10x - (14 - 3x) = 12 \implies 10x - 14 + 3x = 12 \implies 13x - 14 = 12 \]

4. Solve for \(x\):
\[ 13x = 26 \implies x = 2 \]

5. Substitute \(x = 2\) back into the expression for \(y\):
\[ y = \frac{14 - 3(2)}{2} = \frac{14 - 6}{2} = \frac{8}{2} = 4 \]

Thus, the point A is \(A(2, 4)\).

### Step 2: Find the intersection point B of the lines \(4x + 3y = 8\) and \(6x + y = 5\).

Similarly, we solve these two equations simultaneously.

1. From the second equation, express \(y\) in terms of \(x\):
\[ y = 5 - 6x \]

2. Substitute this expression for \(y\) into the first equation:
\[ 4x + 3(5 - 6x) = 8 \]

3. Simplify:
\[ 4x + 15 - 18x = 8 \implies -14x + 15 = 8 \]

4. Solve for \(x\):
\[ -14x = -7 \implies x = \frac{1}{2} \]

5. Substitute \(x = \frac{1}{2}\) back into the expression for \(y\):
\[ y = 5 - 6\left(\frac{1}{2}\right) = 5 - 3 = 2 \]

Thus, the point B is \(B\left(\frac{1}{2}, 2\right)\).

### Step 3: Find the equation of the line AB.

Using the points A(2, 4) and B\(\left(\frac{1}{2}, 2\right)\), we can find the slope \(m\):
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{\frac{1}{2} - 2} = \frac{-2}{-\frac{3}{2}} = \frac{4}{3} \]

Using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \implies y - 4 = \frac{4}{3}(x - 2) \]

Rearranging gives:
\[ y - 4 = \frac{4}{3}x - \frac{8}{3} \implies 4x - 3y + 4 = 0 \]

### Step 4: Calculate the distance from point P(5, -2) to the line AB.

The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Here, \(A = 4\), \(B = -3\), \(C = 4\), and the point \(P(5, -2)\):
\[ d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{16 + 9}} = \frac{|30|}{\sqrt{25}} = \frac{30}{5} = 6 \]

Thus, the distance from point P to line AB is \(6\).

### Final Answer:
The distance of the point \(P(5, -2)\) from the line AB is \(6\).