Find the equation of the line joining the points of intersection of 2x + y=4 with x-y+1=0 and 2x-y-1 =0 and x+y -8=0 :

Options

A. 2x+3y +6 =0
B. 3x+2y +12=0
C. 3x-2y +1=0
D. none of (a) , (b) ,(c)

Answer

To find the equation of the line joining the points of intersection of the equations \(2x + y = 4\), \(x - y + 1 = 0\), \(2x - y - 1 = 0\), and \(x + y - 8 = 0\), we will follow these steps:

### Step 1: Find the points of intersection of the first two lines.
We have the equations:
1. \(2x + y = 4\) (Equation 1)
2. \(x - y + 1 = 0\) (Equation 2)

From Equation 2, we can express \(y\) in terms of \(x\):
\[ y = x + 1 \]

Now, substitute \(y\) in Equation 1:
\[ 2x + (x + 1) = 4 \]
\[ 2x + x + 1 = 4 \]
\[ 3x + 1 = 4 \]
\[ 3x = 3 \implies x = 1 \]

Now substitute \(x = 1\) back into Equation 2 to find \(y\):
\[ y = 1 + 1 = 2 \]

Thus, the first point of intersection is \((1, 2)\).

### Step 2: Find the points of intersection of the next two lines.
Now we will find the intersection of the equations:
1. \(2x - y - 1 = 0\) (Equation 3)
2. \(x + y - 8 = 0\) (Equation 4)

From Equation 4, we can express \(y\) in terms of \(x\):
\[ y = 8 - x \]

Substituting \(y\) in Equation 3:
\[ 2x - (8 - x) - 1 = 0 \]
\[ 2x - 8 + x - 1 = 0 \]
\[ 3x - 9 = 0 \]
\[ 3x = 9 \implies x = 3 \]

Now substitute \(x = 3\) back into Equation 4 to find \(y\):
\[ y = 8 - 3 = 5 \]

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

### Step 3: Find the equation of the line joining the two points \((1, 2)\) and \((3, 5)\).
We can use the two-point form of the equation of a line:
\[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \]
Substituting \((x_1, y_1) = (1, 2)\) and \((x_2, y_2) = (3, 5)\):
\[ \frac{y - 2}{5 - 2} = \frac{x - 1}{3 - 1} \]
\[ \frac{y - 2}{3} = \frac{x - 1}{2} \]

Cross-multiplying gives:
\[ 2(y - 2) = 3(x - 1) \]
\[ 2y - 4 = 3x - 3 \]
Rearranging gives:
\[ 3x - 2y + 1 = 0 \]

### Final Result
The equation of the line joining the points of intersection is:
\[ 3x - 2y + 1 = 0 \]