Given: Three points in a Cartesian plane: $A(5, -2)$, $B(6, 4)$, and $C(7, -2)$.

To Find: Determine whether the triangle formed by these vertices is an isosceles triangle.

Definition: An isosceles triangle is a triangle in which at least two sides are equal in length.

Formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the Distance Formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 1: Calculate the length of side $AB$
Let $A = (5, -2)$ and $B = (6, 4)$.
$AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2}$
$AB = \sqrt{(1)^2 + (4 + 2)^2}$
$AB = \sqrt{1^2 + 6^2}$
$AB = \sqrt{1 + 36}$
$AB = \sqrt{37}$ units

Step 2: Calculate the length of side $BC$
Let $B = (6, 4)$ and $C = (7, -2)$.
$BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2}$
$BC = \sqrt{(1)^2 + (-6)^2}$
$BC = \sqrt{1 + 36}$
$BC = \sqrt{37}$ units

Step 3: Calculate the length of side $AC$
Let $A = (5, -2)$ and $C = (7, -2)$.
$AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2}$
$AC = \sqrt{(2)^2 + (-2 + 2)^2}$
$AC = \sqrt{2^2 + 0^2}$
$AC = \sqrt{4}$
$AC = 2$ units

Step 4: Comparison and Conclusion
We have found the lengths of the sides to be:
$AB = \sqrt{37}$
$BC = \sqrt{37}$
$AC = 2$
[Since $AB = BC = \sqrt{37}$, two sides of the triangle are equal in length.]

Final Answer: Yes, the points (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle because two of its sides are equal in length.

Given: Three points in a Cartesian plane: $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$.

To Find: Determine whether the points $A$, $B$, and $C$ are collinear.

Conceptual Framework: Three points are said to be collinear if they lie on the same straight line. For three points $A$, $B$, and $C$, they are collinear if and only if the sum of the lengths of two smaller segments equals the length of the longest segment (e.g., $AB + BC = AC$, or any permutation thereof).

Formula Used: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the Distance Formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 1: Calculate the distance between points $A(1, 5)$ and $B(2, 3)$

Let $(x_1, y_1) = (1, 5)$ and $(x_2, y_2) = (2, 3)$.

$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2}$

$AB = \sqrt{(1)^2 + (-2)^2}$

$AB = \sqrt{1 + 4}$

$AB = \sqrt{5}$ units

Step 2: Calculate the distance between points $B(2, 3)$ and $C(-2, -11)$

Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (-2, -11)$.

$BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$

$BC = \sqrt{(-4)^2 + (-14)^2}$

$BC = \sqrt{16 + 196}$

$BC = \sqrt{212}$

$BC = \sqrt{4 \times 53} = 2\sqrt{53}$ units

Step 3: Calculate the distance between points $A(1, 5)$ and $C(-2, -11)$

Let $(x_1, y_1) = (1, 5)$ and $(x_2, y_2) = (-2, -11)$.

$AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2}$

$AC = \sqrt{(-3)^2 + (-16)^2}$

$AC = \sqrt{9 + 256}$

$AC = \sqrt{265}$ units

Step 4: Verify the Collinearity Condition

We check if the sum of any two distances equals the third:

$AB + BC = \sqrt{5} + 2\sqrt{53} \approx 2.236 + 14.56 = 16.796$

$AC = \sqrt{265} \approx 16.279$

Since $AB + BC \neq AC$, $AB + AC \neq BC$, and $BC + AC \neq AB$, the condition for collinearity is not satisfied.

Final Answer: The points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are not collinear.

Given: Two points in a Cartesian plane, $P(x_1, y_1) = (0, 0)$ and $Q(x_2, y_2) = (36, 15)$.

To Find: The distance between points $P$ and $Q$, and subsequently, the distance between two towns $A$ and $B$ represented by these coordinates.

Step 1: State the Distance Formula
The distance $d$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$ in a coordinate plane is given by the Euclidean distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
[This formula is derived from the Pythagorean Theorem applied to the right-angled triangle formed by the horizontal and vertical differences of the coordinates.]

Step 2: Substitute the Given Values
Assign the coordinates:
$x_1 = 0, y_1 = 0$
$x_2 = 36, y_2 = 15$
Substituting these into the formula:
$d = \sqrt{(36 - 0)^2 + (15 - 0)^2}$

Step 3: Perform Algebraic Simplification
Calculate the squares of the differences:
$d = \sqrt{(36)^2 + (15)^2}$
Calculate $36^2$: $36 \times 36 = 1296$
Calculate $15^2$: $15 \times 15 = 225$
Sum the squares:
$d = \sqrt{1296 + 225}$
$d = \sqrt{1521}$

Step 4: Extract the Square Root
To find $\sqrt{1521}$, we look for a number which, when multiplied by itself, equals $1521$.
Since $30^2 = 900$ and $40^2 = 1600$, the value must be between 30 and 40.
Testing $39$: $39 \times 39 = 1521$.
$d = 39$

Step 5: Application to Towns A and B
If town $A$ is located at $(0, 0)$ and town $B$ is located at $(36, 15)$, the distance between them is equivalent to the distance calculated between points $P$ and $Q$.
Therefore, the distance between town $A$ and town $B$ is $39$ units.

Final Answer: The distance between the points (0, 0) and (36, 15) is 39 units. Consequently, the distance between town A and town B is 39 units.

Given: Two points in a Cartesian plane, $P(-5, 7)$ and $Q(-1, 3)$.

To find: The distance between points $P$ and $Q$, denoted as $PQ$.

Step 1: Identifying the Distance Formula
To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the Distance Formula derived from the Pythagorean Theorem:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Assigning Coordinates
Let the coordinates of point $P$ be $(x_1, y_1) = (-5, 7)$.
Let the coordinates of point $Q$ be $(x_2, y_2) = (-1, 3)$.

Step 3: Substituting Values into the Formula
Substitute the values of $x_1, y_1, x_2,$ and $y_2$ into the distance formula:

$PQ = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$

Step 4: Performing Arithmetic Operations
First, simplify the expressions inside the parentheses:

$PQ = \sqrt{(-1 + 5)^2 + (-4)^2}$ [Since subtracting a negative is equivalent to addition]

$PQ = \sqrt{(4)^2 + (-4)^2}$

Step 5: Calculating Squares
Calculate the squares of the numbers:

$PQ = \sqrt{16 + 16}$ [Since $4^2 = 16$ and $(-4)^2 = 16$]

$PQ = \sqrt{32}$

Step 6: Simplifying the Radical
Express $\sqrt{32}$ in its simplest form by finding the prime factors:

$32 = 16 \times 2$

$PQ = \sqrt{16 \times 2}$

$PQ = 4\sqrt{2}$ [Taking the square root of 16]

Final Answer: The distance between the points (-5, 7) and (-1, 3) is $4\sqrt{2}$ units.

Given: The coordinates of four points in a Cartesian plane: $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$, and $D(-1, -4)$.

To Find: The type of quadrilateral formed by these points, if any, and the justification for the classification.

Step 1: Distance Formula Application
To determine the nature of the quadrilateral, we calculate the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$) using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step 2: Calculating Side Lengths
For $A(-3, 5)$ and $B(3, 1)$:
$AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{(6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.

For $B(3, 1)$ and $C(0, 3)$:
$BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}$ units.

For $C(0, 3)$ and $D(-1, -4)$:
$CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$ units.

For $D(-1, -4)$ and $A(-3, 5)$:
$DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + (9)^2} = \sqrt{4 + 81} = \sqrt{85}$ units.

Step 3: Analyzing Collinearity
Let us check if the points are collinear or if they form a simple polygon. We observe the slopes of the segments to see if any three points lie on the same line. The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Slope of $AC = \frac{3 - 5}{0 - (-3)} = \frac{-2}{3}$.
Slope of $CB = \frac{1 - 3}{3 - 0} = \frac{-2}{3}$.

[Since the slope of $AC$ is equal to the slope of $CB$ and they share a common point $C$, the points $A, C,$ and $B$ are collinear.]

Step 4: Conclusion
Because points $A, C,$ and $B$ lie on the same straight line, they do not form a quadrilateral. A quadrilateral requires four non-collinear points where no three points are collinear.

Final Answer: The points A, B, C, and D do not form a quadrilateral because the points A, C, and B are collinear.

Given: Two points in a Cartesian plane, $P(x_1, y_1) = (2, 3)$ and $Q(x_2, y_2) = (4, 1)$.

To find: The distance between the points $P$ and $Q$, denoted as $d(P, Q)$.

Step 1: State the Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a coordinate plane is given by the Euclidean distance formula derived from the Pythagorean Theorem:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Identify the coordinates
From the given points:
$x_1 = 2, y_1 = 3$
$x_2 = 4, y_2 = 1$

Step 3: Substitute the values into the formula
$d = \sqrt{(4 - 2)^2 + (1 - 3)^2}$

Step 4: Perform the arithmetic operations inside the parentheses
$d = \sqrt{(2)^2 + (-2)^2}$

Step 5: Calculate the squares
Since $(2)^2 = 4$ and $(-2)^2 = 4$ [Because the square of any real number is non-negative]:
$d = \sqrt{4 + 4}$

Step 6: Simplify the expression
$d = \sqrt{8}$

Step 7: Express in simplest radical form
$d = \sqrt{4 \times 2}$
$d = 2\sqrt{2}$

Final Answer: The distance between the points (2, 3) and (4, 1) is $2\sqrt{2}$ units.

Given: The coordinates of the four points are $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$.

To Find: The type of quadrilateral formed by these points, if any, and the reasons for the classification.

Step 1: Formula Definition
To determine the nature of the quadrilateral, we use the Distance Formula to find the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$).
The Distance Formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Calculating Side Lengths
- Length of AB: $\sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of BC: $\sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
- Length of CD: $\sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of DA: $\sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Step 3: Calculating Diagonal Lengths
- Length of AC: $\sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$ units.
- Length of BD: $\sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.

Step 4: Analysis and Conclusion
From the calculations:
1. Opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = 3\sqrt{2}$.
2. Diagonals are unequal: $AC = 2$ and $BD = \sqrt{52}$.
Since the opposite sides are equal and the diagonals are not equal, the quadrilateral satisfies the properties of a parallelogram.

Final Answer: The points form a parallelogram.

Given: The positions of four friends seated in a classroom are represented by coordinates on a Cartesian plane. Based on the provided figure, the coordinates are:

$A = (3, 4)$

$B = (6, 7)$

$C = (9, 4)$

$D = (6, 1)$

To Find: Determine whether the quadrilateral $ABCD$ is a square by calculating the lengths of its sides and diagonals using the distance formula.

Visual Representation:

Step 1: The Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Calculating Side Lengths
We calculate the lengths of the four sides $AB, BC, CD,$ and $DA$.

Length of $AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of $BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of $CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of $DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

[Since all sides are equal ($AB = BC = CD = DA = 3\sqrt{2}$), the quadrilateral is at least a rhombus.]

Step 3: Calculating Diagonal Lengths
For a rhombus to be a square, its diagonals must also be equal in length.

Length of diagonal $AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{(6)^2 + (0)^2} = \sqrt{36} = 6$ units.

Length of diagonal $BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{(0)^2 + (-6)^2} = \sqrt{36} = 6$ units.

Step 4: Conclusion
Since all four sides are equal ($3\sqrt{2}$ units) and both diagonals are equal ($6$ units), the quadrilateral $ABCD$ satisfies the geometric properties of a square.

Final Answer: Champa is correct; ABCD is a square.

Given: Two points $A(2, -5)$ and $B(-2, 9)$. A point $P$ lies on the $x$-axis such that it is equidistant from $A$ and $B$.

To Find: The coordinates of point $P$.

Step 1: Defining the coordinates of point P
Since point $P$ lies on the $x$-axis, its $y$-coordinate must be $0$. Let the coordinates of point $P$ be $(x, 0)$.

Step 2: Applying the Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 3: Setting up the Equidistance Equation
Since $P$ is equidistant from $A(2, -5)$ and $B(-2, 9)$, we have $PA = PB$.
Therefore, $PA^2 = PB^2$ [Squaring both sides to eliminate the square root].

Step 4: Calculating the squared distances
For $PA^2$:
$PA^2 = (x - 2)^2 + (0 - (-5))^2$
$PA^2 = (x - 2)^2 + (5)^2$
$PA^2 = x^2 - 4x + 4 + 25$
$PA^2 = x^2 - 4x + 29$

For $PB^2$:
$PB^2 = (x - (-2))^2 + (0 - 9)^2$
$PB^2 = (x + 2)^2 + (-9)^2$
$PB^2 = x^2 + 4x + 4 + 81$
$PB^2 = x^2 + 4x + 85$

Step 5: Solving for x
Equating $PA^2 = PB^2$:
$x^2 - 4x + 29 = x^2 + 4x + 85$
Subtract $x^2$ from both sides:
$-4x + 29 = 4x + 85$
Subtract $4x$ from both sides:
$-8x + 29 = 85$
Subtract $29$ from both sides:
$-8x = 85 - 29$
$-8x = 56$
Divide by $-8$:
$x = \frac{56}{-8}$
$x = -7$

Final Answer: The point on the x-axis equidistant from (2, –5) and (–2, 9) is (-7, 0).

Given:

Point $Q(0, 1)$ is equidistant from points $P(5, -3)$ and $R(x, 6)$.

To find:

1. The value(s) of $x$.

2. The distance $QR$.

3. The distance $PR$.

Step 1: Applying the Distance Formula

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Since $Q$ is equidistant from $P$ and $R$, we have $QP = QR$. Therefore, $QP^2 = QR^2$.

Step 2: Calculating $QP^2$

For $Q(0, 1)$ and $P(5, -3)$:

$QP^2 = (5 - 0)^2 + (-3 - 1)^2$

$QP^2 = (5)^2 + (-4)^2$

$QP^2 = 25 + 16 = 41$

Step 3: Calculating $QR^2$ and solving for $x$

For $Q(0, 1)$ and $R(x, 6)$:

$QR^2 = (x - 0)^2 + (6 - 1)^2$

$QR^2 = x^2 + (5)^2$

$QR^2 = x^2 + 25$

Equating $QP^2 = QR^2$:

$41 = x^2 + 25$

$x^2 = 41 - 25$

$x^2 = 16$

$x = \pm \sqrt{16} = \pm 4$

Thus, the possible coordinates for $R$ are $(4, 6)$ or $(-4, 6)$.

Step 4: Finding the distance $QR$

Since $QP = QR$ and $QP^2 = 41$, then $QR = \sqrt{41}$.

Step 5: Finding the distance $PR$

Case 1: If $R = (4, 6)$ and $P = (5, -3)$:

$PR = \sqrt{(4 - 5)^2 + (6 - (-3))^2}$

$PR = \sqrt{(-1)^2 + (9)^2} = \sqrt{1 + 81} = \sqrt{82}$

Case 2: If $R = (-4, 6)$ and $P = (5, -3)$:

$PR = \sqrt{(-4 - 5)^2 + (6 - (-3))^2}$

$PR = \sqrt{(-9)^2 + (9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$

Final Answer:

The values of $x$ are $4$ or $-4$. The distance $QR = \sqrt{41}$. The distance $PR$ is $\sqrt{82}$ (when $x=4$) or $9\sqrt{2}$ (when $x=-4$).

Given:

The coordinates of point $P$ are $(x_1, y_1) = (2, -3)$.

The coordinates of point $Q$ are $(x_2, y_2) = (10, y)$.

The distance between points $P$ and $Q$ is $d = 10$ units.

To find:

The value(s) of $y$.

Step 1: Applying the Distance Formula

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian plane is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

[By the Distance Formula in Coordinate Geometry]

Step 2: Substituting the Given Values

Substitute $d = 10$, $x_1 = 2$, $y_1 = -3$, $x_2 = 10$, and $y_2 = y$ into the formula:

$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$

$10 = \sqrt{(8)^2 + (y + 3)^2}$

Step 3: Squaring Both Sides to Eliminate the Square Root

To solve for $y$, we square both sides of the equation:

$(10)^2 = (\sqrt{(8)^2 + (y + 3)^2})^2$

$100 = 8^2 + (y + 3)^2$

$100 = 64 + (y + 3)^2$

Step 4: Simplifying the Equation

Subtract 64 from both sides:

$100 - 64 = (y + 3)^2$

$36 = (y + 3)^2$

Step 5: Solving for $y$

Take the square root of both sides, remembering to include both positive and negative roots:

$\pm \sqrt{36} = y + 3$

$\pm 6 = y + 3$

This yields two distinct linear equations:

Case 1: $6 = y + 3$

$y = 6 - 3$

$y = 3$

Case 2: $-6 = y + 3$

$y = -6 - 3$

$y = -9$

Conclusion:

The possible values for $y$ that satisfy the condition are $3$ and $-9$.

Final Answer: The values of $y$ are $3$ and $-9$.

Given: Two points in a Cartesian plane, $P(x_1, y_1) = (a, b)$ and $Q(x_2, y_2) = (-a, -b)$.

To find: The distance between the points $P$ and $Q$.

Visual Representation:

Step 1: State the Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the Euclidean distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
[This formula is derived from the Pythagorean Theorem applied to the horizontal and vertical differences between coordinates.]

Step 2: Substitute the given coordinates into the formula
Given $x_1 = a$, $y_1 = b$, $x_2 = -a$, and $y_2 = -b$.
Substituting these values into the distance formula:
$d = \sqrt{(-a - a)^2 + (-b - b)^2}$

Step 3: Simplify the expressions inside the parentheses
Perform the subtraction within the brackets:
$(-a - a) = -2a$
$(-b - b) = -2b$
Therefore:
$d = \sqrt{(-2a)^2 + (-2b)^2}$

Step 4: Evaluate the squares
Recall that $(-x)^2 = x^2$:
$(-2a)^2 = 4a^2$
$(-2b)^2 = 4b^2$
Substituting these back into the equation:
$d = \sqrt{4a^2 + 4b^2}$

Step 5: Factor out the common term and simplify the radical
Factor out $4$ from the terms inside the square root:
$d = \sqrt{4(a^2 + b^2)}$
Using the property of radicals $\sqrt{xy} = \sqrt{x} \cdot \sqrt{y}$:
$d = \sqrt{4} \cdot \sqrt{a^2 + b^2}$
Since $\sqrt{4} = 2$:
$d = 2\sqrt{a^2 + b^2}$

Final Answer: The distance between the points $(a, b)$ and $(-a, -b)$ is $2\sqrt{a^2 + b^2}$ units.

Given: The coordinates of the four points forming a quadrilateral are $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, and $D(-3, 0)$.

To Find: The type of quadrilateral formed by these points, with supporting reasons.

Step 1: Distance Formula Application
To determine the type of quadrilateral, we calculate the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$) using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step 2: Calculating Side Lengths
For $A(-1, -2)$ and $B(1, 0)$:
$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $B(1, 0)$ and $C(-1, 2)$:
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $C(-1, 2)$ and $D(-3, 0)$:
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $D(-3, 0)$ and $A(-1, -2)$:
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

Step 3: Calculating Diagonal Lengths
For $A(-1, -2)$ and $C(-1, 2)$:
$AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.

For $B(1, 0)$ and $D(-3, 0)$:
$BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.

Step 4: Conclusion based on Properties
Since all four sides are equal ($AB = BC = CD = DA = 2\sqrt{2}$) and both diagonals are equal ($AC = BD = 4$), the quadrilateral satisfies the necessary and sufficient conditions for a square.

Final Answer: The quadrilateral formed is a square.

Given: A point $P(x, y)$ is equidistant from point $A(3, 6)$ and point $B(-3, 4)$.

To Find: A relation between $x$ and $y$.

Step 1: Understanding the Condition of Equidistance
Since point $P(x, y)$ is equidistant from $A(3, 6)$ and $B(-3, 4)$, the distance $PA$ must be equal to the distance $PB$. Mathematically, $PA = PB$, which implies $PA^2 = PB^2$.

Step 2: Applying the Distance Formula
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Therefore, the square of the distance is:
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Step 3: Setting up the Equation
For $PA^2$:
$PA^2 = (x - 3)^2 + (y - 6)^2$
For $PB^2$:
$PB^2 = (x - (-3))^2 + (y - 4)^2 = (x + 3)^2 + (y - 4)^2$

Step 4: Expanding the Algebraic Expressions
Using the identity $(a - b)^2 = a^2 - 2ab + b^2$ and $(a + b)^2 = a^2 + 2ab + b^2$:
$PA^2 = (x^2 - 6x + 9) + (y^2 - 12y + 36)$
$PB^2 = (x^2 + 6x + 9) + (y^2 - 8y + 16)$

Step 5: Equating and Simplifying
Since $PA^2 = PB^2$:
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
Subtract $x^2$ and $y^2$ from both sides:
$-6x - 12y + 45 = 6x - 8y + 25$
Rearrange the terms to one side:
$-6x - 6x - 12y + 8y + 45 - 25 = 0$
$-12x - 4y + 20 = 0$

Step 6: Final Simplification
Divide the entire equation by $-4$ to simplify:
$3x + y - 5 = 0$
Or, $3x + y = 5$.

Final Answer: 3x + y = 5