Prove that If vertices of a `Delta` are (2,3) (5,-4) (1,-3) The triangle can't be formed

Answer

To prove that a triangle cannot be formed with the vertices (2, 3), (5, -4), and (1, -3), we will calculate the lengths of the sides of the triangle formed by these vertices and then check the triangle inequality theorem.

### Step 1: Calculate the lengths of the sides of the triangle.

1. **Finding the length of side AB** (between points A(2, 3) and B(5, -4)):
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5 - 2)^2 + (-4 - 3)^2} \]
\[ = \sqrt{(3)^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58} \]

2. **Finding the length of side BC** (between points B(5, -4) and C(1, -3)):
\[ BC = \sqrt{(1 - 5)^2 + (-3 + 4)^2} = \sqrt{(-4)^2 + (1)^2} \]
\[ = \sqrt{16 + 1} = \sqrt{17} \]

3. **Finding the length of side CA** (between points C(1, -3) and A(2, 3)):
\[ CA = \sqrt{(2 - 1)^2 + (3 + 3)^2} = \sqrt{(1)^2 + (6)^2} \]
\[ = \sqrt{1 + 36} = \sqrt{37} \]

### Step 2: Check the triangle inequality theorem.

The triangle inequality states that for any triangle with sides of lengths a, b, and c:
- \( a + b > c \)
- \( a + c > b \)
- \( b + c > a \)

Let:
- \( a = \sqrt{58} \)
- \( b = \sqrt{17} \)
- \( c = \sqrt{37} \)

Now we will check the inequalities:

1. **Check if \( b + c > a \)**:
\[ \sqrt{17} + \sqrt{37} > \sqrt{58} \]
Squaring both sides:
\[ 17 + 37 + 2\sqrt{17 \cdot 37} > 58 \]
\[ 54 + 2\sqrt{629} > 58 \]
This is true because \( 2\sqrt{629} \) is a positive number.

2. **Check if \( a + c > b \)**:
\[ \sqrt{58} + \sqrt{37} > \sqrt{17} \]
Squaring both sides:
\[ 58 + 37 + 2\sqrt{58 \cdot 37} > 17 \]
\[ 95 + 2\sqrt{2146} > 17 \]
This is also true.

3. **Check if \( a + b > c \)**:
\[ \sqrt{58} + \sqrt{17} > \sqrt{37} \]
Squaring both sides:
\[ 58 + 17 + 2\sqrt{58 \cdot 17} > 37 \]
\[ 75 + 2\sqrt{986} > 37 \]
This is true as well.

### Conclusion:
Since all three inequalities are satisfied, it appears that a triangle can be formed with the given vertices. However, we need to check if the points are collinear.

### Step 3: Check if the points are collinear.

To check if the points are collinear, we can calculate the area of the triangle formed by these points using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 2(-4 + 3) + 5(-3 - 3) + 1(3 + 4) \right| \]
\[ = \frac{1}{2} \left| 2(-1) + 5(-6) + 1(7) \right| \]
\[ = \frac{1}{2} \left| -2 - 30 + 7 \right| = \frac{1}{2} \left| -25 \right| = \frac{25}{2} \]
Since the area is not equal to zero, the points are not collinear.

### Final Conclusion:
Thus, the triangle can be formed with the given vertices (2, 3), (5, -4), and (1, -3).