3. Find the coordinates of a vertex $$ D $$ of the equadriateral $$ A(0,4) B(-4,7) $$ $$ (-7,3) $$ and $$ \Delta $$ if $$ A D=B C $$ and $$ D $$ is on $$ x $$-akis.

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To determine the coordinates of vertex $$ D $$ in the quadrilateral $$ ABCD $$ with the given conditions, let's break down the problem step by step.

### Given:
- **Vertices:**
  - $$ A(0, 4) $$
  - $$ B(-4, 7) $$
  - $$ C(-7, 3) $$
- **Conditions:**
  - $$ AD = BC $$
  - $$ D $$ lies on the $$ x $$-axis.

### Objective:
Find the coordinates of vertex $$ D $$.

### Step 1: Calculate the Distance $$ BC $$
First, find the distance between points $$ B $$ and $$ C $$.

$$
B(-4, 7) \quad \text{and} \quad C(-7, 3)
$$

Using the distance formula:
$$
BC = \sqrt{(-7 - (-4))^2 + (3 - 7)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
$$

So, $$ BC = 5 $$ units.

### Step 2: Set Up the Distance $$ AD $$
Since $$ AD = BC $$, we have $$ AD = 5 $$ units.

Point $$ D $$ lies on the $$ x $$-axis, so its coordinates are $$ (x, 0) $$.

$$
A(0, 4) \quad \text{and} \quad D(x, 0)
$$

Using the distance formula:
$$
AD = \sqrt{(x - 0)^2 + (0 - 4)^2} = \sqrt{x^2 + 16} = 5
$$

### Step 3: Solve for $$ x $$
Set the distance equal to 5 and solve for $$ x $$:

$$
\sqrt{x^2 + 16} = 5
$$
$$
x^2 + 16 = 25
$$
$$
x^2 = 9
$$
$$
x = \pm 3
$$

### Step 4: Determine the Coordinates of $$ D $$
Thus, there are two possible coordinates for $$ D $$:

1. $$ D(3, 0) $$
2. $$ D(-3, 0) $$

### Conclusion
Both points satisfy the given conditions. Therefore, the coordinates of vertex $$ D $$ can be either $$ (3, 0) $$ or $$ (-3, 0) $$.

**Final Answer:**
The vertex D lies on the x‑axis at ( 3 , 0 ) or at ( –3, 0 ).
The coordinates of vertex $$ D $$ are either $$ (3, 0) $$ or $$ (-3, 0) $$.

Find the coordinates of a vertex
of the equadriateral
and if and
is on -akis.

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Find the coordinates of a vertex of the equadriateral and if and is on -akis.