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:

1. The school ground is rectangular, represented by a coordinate plane where the x-axis represents the lines (1 to 10) and the y-axis represents the distance along AD (in meters).

2. Total distance along AD = $100$ meters (since there are 100 flower pots placed 1m apart).

3. Niharika runs on the 2nd line ($x_1 = 2$) and covers $\frac{1}{4}$ of the distance AD.

4. Preet runs on the 8th line ($x_2 = 8$) and covers $\frac{1}{5}$ of the distance AD.

To Find:

1. The distance between the green flag (Niharika) and the red flag (Preet).

2. The coordinates where Rashmi should post her blue flag (the midpoint of the two flags).

Step 1: Determine the coordinates of the flags.

Let the position of the green flag be $G(x_1, y_1)$.

$x_1 = 2$

$y_1 = \frac{1}{4} \times 100 = 25$

So, $G = (2, 25)$.

Let the position of the red flag be $R(x_2, y_2)$.

$x_2 = 8$

$y_2 = \frac{1}{5} \times 100 = 20$

So, $R = (8, 20)$.

Step 2: Calculate the distance between the two flags.

Using the Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(8 - 2)^2 + (20 - 25)^2}$

$d = \sqrt{(6)^2 + (-5)^2}$

$d = \sqrt{36 + 25}$

$d = \sqrt{61}$

$d \approx 7.81$ meters.

Step 3: Determine the position of the blue flag.

The blue flag is at the midpoint $M(x, y)$ of the line segment joining $G(2, 25)$ and $R(8, 20)$.

Using the Midpoint Formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$

$y = \frac{25 + 20}{2} = \frac{45}{2} = 22.5$

Therefore, the blue flag should be posted on the 5th line at a distance of 22.5 meters from AD.

Final Answer: The distance between the two flags is $\sqrt{61}$ m (approx 7.81 m). Rashmi should post her blue flag at the coordinates (5, 22.5).

Given:
The coordinates of two points are $A(x_1, y_1) = (-1, 7)$ and $B(x_2, y_2) = (4, -3)$.
The ratio in which the point divides the line segment $AB$ is $m : n = 2 : 3$.

To Find:
The coordinates of the point $P(x, y)$ that divides the line segment joining $A$ and $B$ in the ratio $2 : 3$.

Step 1: Applying the Section Formula
The Section Formula states that the coordinates of a point $P(x, y)$ dividing the line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m : n$ are given by:
$x = \frac{mx_2 + nx_1}{m + n}$
$y = \frac{my_2 + ny_1}{m + n}$

Step 2: Identifying the variables
From the given information:
$x_1 = -1, y_1 = 7$
$x_2 = 4, y_2 = -3$
$m = 2, n = 3$

Step 3: Calculating the x-coordinate
Substituting the values into the formula for $x$:
$x = \frac{(2)(4) + (3)(-1)}{2 + 3}$
$x = \frac{8 - 3}{5}$ [Performing multiplication and addition]
$x = \frac{5}{5}$
$x = 1$

Step 4: Calculating the y-coordinate
Substituting the values into the formula for $y$:
$y = \frac{(2)(-3) + (3)(7)}{2 + 3}$
$y = \frac{-6 + 21}{5}$ [Performing multiplication and addition]
$y = \frac{15}{5}$
$y = 3$

Final Answer:
The coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3 are (1, 3).

Given:

1. A circle with center $O$ at coordinates $(2, -3)$.

2. $AB$ is the diameter of the circle.

3. The coordinates of point $B$ are $(1, 4)$.

To Find:

The coordinates of point $A$, let us denote them as $(x, y)$.

Step 1: Theoretical Basis

In a circle, the center $O$ is the midpoint of any diameter $AB$. According to the Midpoint Formula, if a line segment has endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint $(x_m, y_m)$ is given by:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

Step 2: Assigning Variables

Let the coordinates of point $A$ be $(x, y)$.

Let the coordinates of point $B$ be $(x_2, y_2) = (1, 4)$.

Let the coordinates of the center $O$ be $(x_m, y_m) = (2, -3)$.

Step 3: Calculating the x-coordinate of A

Using the midpoint formula for the x-coordinate:

$2 = \frac{x + 1}{2}$

[Multiplying both sides by 2]

$4 = x + 1$

[Subtracting 1 from both sides]

$x = 4 - 1$

$x = 3$

Step 4: Calculating the y-coordinate of A

Using the midpoint formula for the y-coordinate:

$-3 = \frac{y + 4}{2}$

[Multiplying both sides by 2]

$-6 = y + 4$

[Subtracting 4 from both sides]

$y = -6 - 4$

$y = -10$

Final Answer: The coordinates of point A are (3, -10).

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.