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

Options

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

Answer

To solve the problem step by step, we need to find the coordinates of points P and Q based on the given equations, and then derive the equation of the line PQ.

### Step 1: Find the coordinates of point P(a, b)

Point P lies on the line given by the equation:
\[ 3x + 2y = 13 \]

Substituting the coordinates of point P into the equation, we have:
\[ 3a + 2b = 13 \quad \text{(1)} \]

### Step 2: Find the coordinates of point Q(b, a)

Point Q lies on the line given by the equation:
\[ 4x - y = 5 \]

Substituting the coordinates of point Q into the equation, we have:
\[ 4b - a = 5 \quad \text{(2)} \]

### Step 3: Solve the system of equations

We now have a system of two equations:
1. \( 3a + 2b = 13 \) (1)
2. \( 4b - a = 5 \) (2)

From equation (2), we can express \( a \) in terms of \( b \):
\[ a = 4b - 5 \]

### Step 4: Substitute \( a \) into equation (1)

Now substitute \( a \) from equation (2) into equation (1):
\[ 3(4b - 5) + 2b = 13 \]
\[ 12b - 15 + 2b = 13 \]
\[ 14b - 15 = 13 \]
\[ 14b = 28 \]
\[ b = 2 \]

### Step 5: Find the value of \( a \)

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

### Step 6: Determine the coordinates of points P and Q

Now we have:
- Point P: \( (a, b) = (3, 2) \)
- Point Q: \( (b, a) = (2, 3) \)

### Step 7: Find the equation of line PQ

Using the two-point form of the equation of a line, which is:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]

Here, \( (x_1, y_1) = (3, 2) \) and \( (x_2, y_2) = (2, 3) \):
\[ y - 2 = \frac{3 - 2}{2 - 3}(x - 3) \]
\[ y - 2 = \frac{1}{-1}(x - 3) \]
\[ y - 2 = - (x - 3) \]
\[ y - 2 = -x + 3 \]
\[ y + x - 5 = 0 \]

Thus, the equation of line PQ is:
\[ x + y = 5 \]

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