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:

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 $x=-4$).