The point P(a,b) and Q(b,a) lie on the lines `3x+2y-13=0` and `4x-y-5=0`. The equation of line PQ is

Options

A. `x-y=5`
B. `x+y=5`
C. `x-y=-5`
D. `x+y=-5`

Answer

To find the equation of the line PQ, we will follow these steps:

### Step 1: Set up the equations for points P and Q
Given that point P(a, b) lies on the line \(3x + 2y - 13 = 0\), we can substitute the coordinates of P into the equation:
\[ 3a + 2b - 13 = 0 \quad \text{(Equation 1)} \]
Similarly, point Q(b, a) lies on the line \(4x - y - 5 = 0\), so we substitute the coordinates of Q:
\[ 4b - a - 5 = 0 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations simultaneously
From Equation 1:
\[ 3a + 2b = 13 \quad \text{(Rearranging Equation 1)} \]

From Equation 2:
\[ 4b - a = 5 \quad \text{(Rearranging Equation 2)} \]

Now, we can express \(a\) from Equation 2:
\[ a = 4b - 5 \quad \text{(Substituting for a)} \]

### Step 3: Substitute \(a\) into Equation 1
Substituting \(a = 4b - 5\) into \(3a + 2b = 13\):
\[ 3(4b - 5) + 2b = 13 \]
Expanding this:
\[ 12b - 15 + 2b = 13 \]
Combining like terms:
\[ 14b - 15 = 13 \]
Adding 15 to both sides:
\[ 14b = 28 \]
Dividing by 14:
\[ b = 2 \]

### Step 4: Find \(a\)
Now substitute \(b = 2\) back into the equation for \(a\):
\[ a = 4(2) - 5 = 8 - 5 = 3 \]

### Step 5: Determine the coordinates of points P and Q
Thus, the coordinates of point P are:
\[ P(3, 2) \]
And the coordinates of point Q are:
\[ Q(2, 3) \]

### Step 6: Find the equation of line PQ
Using the two points P(3, 2) and Q(2, 3), we can use the point-slope form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \(y_1 = 2\), \(y_2 = 3\), \(x_1 = 3\), and \(x_2 = 2\):
\[ y - 2 = \frac{3 - 2}{2 - 3}(x - 3) \]
This simplifies to:
\[ y - 2 = \frac{1}{-1}(x - 3) \]
\[ y - 2 = - (x - 3) \]
\[ y - 2 = -x + 3 \]
Rearranging gives:
\[ x + y = 5 \]

### Final Answer
The equation of line PQ is:
\[ x + y = 5 \]