Find the equation of the straight line which passes through the point of intersection of the straight lines `x+y=8 and 3x-2y+1=0` and is parallel to the straight line joining the points (3, 4) and (5, 6).

Answer

To find the equation of the straight line that passes through the point of intersection of the lines \( x + y = 8 \) and \( 3x - 2y + 1 = 0 \), and is parallel to the line joining the points \( (3, 4) \) and \( (5, 6) \), we will follow these steps:

### Step 1: Find the point of intersection of the two lines.

We have the equations:
1. \( x + y = 8 \) (Equation 1)
2. \( 3x - 2y + 1 = 0 \) (Equation 2)

Rearranging Equation 2 gives us:
\[ 3x - 2y = -1 \]

Now, we will solve these equations simultaneously.

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 8 - x \]

Substituting this into Equation 2:
\[ 3x - 2(8 - x) = -1 \]
\[ 3x - 16 + 2x = -1 \]
\[ 5x - 16 = -1 \]
\[ 5x = 15 \]
\[ x = 3 \]

Now substituting \( x = 3 \) back into Equation 1 to find \( y \):
\[ 3 + y = 8 \]
\[ y = 5 \]

Thus, the point of intersection is \( (3, 5) \).

### Step 2: Calculate the slope of the line joining the points \( (3, 4) \) and \( (5, 6) \).

Using the slope formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]

Let \( (x_1, y_1) = (3, 4) \) and \( (x_2, y_2) = (5, 6) \):
\[ \text{slope} = \frac{6 - 4}{5 - 3} = \frac{2}{2} = 1 \]

### Step 3: Write the equation of the line parallel to the slope found in Step 2.

Since the required line is parallel to the line joining the two points, it will have the same slope of \( 1 \).

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]

Where \( m = 1 \), \( (x_1, y_1) = (3, 5) \):
\[ y - 5 = 1(x - 3) \]
\[ y - 5 = x - 3 \]
\[ y = x + 2 \]

### Step 4: Write the final equation in standard form.

Rearranging the equation:
\[ y - x - 2 = 0 \]

Thus, the required equation of the straight line is:
\[ y - x - 2 = 0 \]