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):

Options

A. (A) x - y + 2 = 0
B. (B) x + y - 2 = 0
C. (C) 3x - 4y + 8 = 0
D. (D) none of these

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 can 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)

To find the point of intersection, we can solve these equations simultaneously.

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

Now, substitute \(y\) in Equation 2:
\[ 3x - 2(8 - x) + 1 = 0 \]
\[ 3x - 16 + 2x + 1 = 0 \]
\[ 5x - 15 = 0 \]
\[ x = 3 \]

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

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

### Step 2: Find the slope of the line joining the points (3, 4) and (5, 6)
The slope \(m\) of the line joining the points \((x_1, y_1) = (3, 4)\) and \((x_2, y_2) = (5, 6)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 4}{5 - 3} = \frac{2}{2} = 1 \]

### Step 3: Write the equation of the line through the point of intersection with the found slope
Since we have a point \((3, 5)\) and a slope \(m = 1\), we can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - 5 = 1(x - 3) \]
\[ y - 5 = x - 3 \]
\[ y = x + 2 \]

### Step 4: Convert the equation into standard form
To convert \(y = x + 2\) into standard form \(Ax + By + C = 0\):
\[ x - y + 2 = 0 \]

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