Find the area of the triangle formed by the straight lines whose equations are 2y+x−5=0,y+2x−7=0, and x−y+1=0.

Options

A. a. `3/2`
B. b. 10
C. c. 6
D. d. `2/5`

Answer

To find the area of the triangle formed by the lines given by the equations \(2y + x - 5 = 0\), \(y + 2x - 7 = 0\), and \(x - y + 1 = 0\), we will first find the points of intersection of these lines, which will serve as the vertices of the triangle.

### Step 1: Find the intersection points of the lines

**1. Intersection of Line 1 and Line 2:**

The equations are:
- Line 1: \(2y + x - 5 = 0\) → \(x + 2y = 5\) (Equation 1)
- Line 2: \(y + 2x - 7 = 0\) → \(2x + y = 7\) (Equation 2)

We can solve these two equations simultaneously. From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = 5 - 2y \]
Now, substitute this expression for \(x\) into Equation 2:
\[ 2(5 - 2y) + y = 7 \]
Expanding this gives:
\[ 10 - 4y + y = 7 \]
Combining like terms:
\[ 10 - 3y = 7 \]
Solving for \(y\):
\[ -3y = 7 - 10 \implies -3y = -3 \implies y = 1 \]
Now substitute \(y = 1\) back into Equation 1 to find \(x\):
\[ x + 2(1) = 5 \implies x + 2 = 5 \implies x = 3 \]
Thus, the intersection point of Line 1 and Line 2 is \((3, 1)\).

**2. Intersection of Line 2 and Line 3:**

The equations are:
- Line 2: \(y + 2x - 7 = 0\) → \(2x + y = 7\) (Equation 2)
- Line 3: \(x - y + 1 = 0\) → \(x - y = -1\) (Equation 3)

From Equation 3, we can express \(y\) in terms of \(x\):
\[ y = x + 1 \]
Substituting this into Equation 2:
\[ 2x + (x + 1) = 7 \]
This simplifies to:
\[ 3x + 1 = 7 \implies 3x = 6 \implies x = 2 \]
Now substitute \(x = 2\) back into Equation 3 to find \(y\):
\[ y = 2 + 1 = 3 \]
Thus, the intersection point of Line 2 and Line 3 is \((2, 3)\).

**3. Intersection of Line 1 and Line 3:**

The equations are:
- Line 1: \(2y + x - 5 = 0\) → \(x + 2y = 5\) (Equation 1)
- Line 3: \(x - y + 1 = 0\) → \(x - y = -1\) (Equation 3)

From Equation 3, we can express \(x\) in terms of \(y\):
\[ x = y - 1 \]
Substituting this into Equation 1:
\[ (y - 1) + 2y = 5 \]
This simplifies to:
\[ 3y - 1 = 5 \implies 3y = 6 \implies y = 2 \]
Now substitute \(y = 2\) back into Equation 3 to find \(x\):
\[ x = 2 - 1 = 1 \]
Thus, the intersection point of Line 1 and Line 3 is \((1, 2)\).

### Step 2: Vertices of the Triangle

The vertices of the triangle formed by the intersection of the lines are:
- \(A(3, 1)\)
- \(B(2, 3)\)
- \(C(1, 2)\)

### Step 3: Calculate the Area of the Triangle

The area \(A\) of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of the vertices:
\[ A = \frac{1}{2} \left| 3(3 - 2) + 2(2 - 1) + 1(1 - 3) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| 3(1) + 2(1) + 1(-2) \right| \]
\[ = \frac{1}{2} \left| 3 + 2 - 2 \right| = \frac{1}{2} \left| 3 \right| = \frac{3}{2} \]

Thus, the area of the triangle is \(\frac{3}{2}\).

### Final Answer
The area of the triangle formed by the given lines is \(\frac{3}{2}\).