Let the vertices of a triangle ABC be `(7,9) , (3,-7)` and (-3,3) then the triangle is :

Options

A. a. right angled
B. b. equilateral
C. c. isosceles
D. d. both (a) and (c)

Answer

To determine the type of triangle formed by the vertices A(7, 9), B(3, -7), and C(-3, 3), we will calculate the lengths of the sides AB, BC, and CA using the distance formula.

### Step 1: Calculate the Length of Side AB
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

For points A(7, 9) and B(3, -7):
- \( x1 = 7, y1 = 9 \)
- \( x2 = 3, y2 = -7 \)

Using the distance formula:
\[ AB = \sqrt{(3 - 7)^2 + (-7 - 9)^2} \]
\[ = \sqrt{(-4)^2 + (-16)^2} \]
\[ = \sqrt{16 + 256} \]
\[ = \sqrt{272} \]
\[ = 4\sqrt{17} \]

### Step 2: Calculate the Length of Side BC
For points B(3, -7) and C(-3, 3):
- \( x1 = 3, y1 = -7 \)
- \( x2 = -3, y2 = 3 \)

Using the distance formula:
\[ BC = \sqrt{(-3 - 3)^2 + (3 - (-7))^2} \]
\[ = \sqrt{(-6)^2 + (10)^2} \]
\[ = \sqrt{36 + 100} \]
\[ = \sqrt{136} \]
\[ = 2\sqrt{34} \]

### Step 3: Calculate the Length of Side CA
For points C(-3, 3) and A(7, 9):
- \( x1 = -3, y1 = 3 \)
- \( x2 = 7, y2 = 9 \)

Using the distance formula:
\[ CA = \sqrt{(7 - (-3))^2 + (9 - 3)^2} \]
\[ = \sqrt{(10)^2 + (6)^2} \]
\[ = \sqrt{100 + 36} \]
\[ = \sqrt{136} \]
\[ = 2\sqrt{34} \]

### Step 4: Compare the Lengths of the Sides
Now we have:
- \( AB = 4\sqrt{17} \)
- \( BC = 2\sqrt{34} \)
- \( CA = 2\sqrt{34} \)

From the calculations:
- \( BC = CA \)
- \( AB \neq BC \) and \( AB \neq CA \)

### Conclusion
Since two sides (BC and CA) are equal and the third side (AB) is not equal to them, triangle ABC is an **isosceles triangle**.