Verify the following: `(0," "7," "10)` , `(1," "6," "6)` and `(" "4," "9," "6)` are the vertices of

Options

A. (a)Right triangle
B. (b)Isosceles triangle
C. (c)lsosceles right triangle
D. (d)Equilateral triangle

Answer

To verify if the points \( A(0, 7, 10) \), \( B(1, 6, 6) \), and \( C(4, 9, 6) \) are the vertices of a right-angled isosceles triangle, we will follow these steps:

### Step 1: Calculate the distances between the points

We will use the distance formula in three-dimensional geometry, which is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

#### Distance \( AB \):

\[ AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} \]
\[ = \sqrt{(1)^2 + (-1)^2 + (-4)^2} \]
\[ = \sqrt{1 + 1 + 16} = \sqrt{18} \]

#### Distance \( BC \):

\[ BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (6 - 6)^2} \]
\[ = \sqrt{(3)^2 + (3)^2 + (0)^2} \]
\[ = \sqrt{9 + 9 + 0} = \sqrt{18} \]

#### Distance \( AC \):

\[ AC = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2} \]
\[ = \sqrt{(4)^2 + (2)^2 + (-4)^2} \]
\[ = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]

### Step 2: Check if it is a right-angled triangle

For the triangle to be a right-angled triangle, the Pythagorean theorem must hold:

\[ AC^2 = AB^2 + BC^2 \]

Calculating the squares:

\[ AC^2 = 6^2 = 36 \]
\[ AB^2 = (\sqrt{18})^2 = 18 \]
\[ BC^2 = (\sqrt{18})^2 = 18 \]

Now checking the Pythagorean theorem:

\[ AC^2 = AB^2 + BC^2 \implies 36 = 18 + 18 \]
\[ 36 = 36 \quad \text{(True)} \]

### Step 3: Check if it is an isosceles triangle

Since \( AB = BC = \sqrt{18} \), the triangle is isosceles because two sides are equal.

### Conclusion

Since the triangle formed by the points \( A(0, 7, 10) \), \( B(1, 6, 6) \), and \( C(4, 9, 6) \) satisfies both the conditions of being a right-angled triangle and being isosceles, we can conclude that:

**The given vertices are the vertices of a right-angled isosceles triangle.**