Question 9

Equidistant: A point is equidistant from two other points when its distance from both is exactly equal. So here, $QP = QR$.

Distance Formula: The distance between two points $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$:

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

We set $QP = QR$, substitute the coordinates, and solve for $x$.

We will use the distance formula to find the unknown coordinate and distances.

Step 1 — Find the values of x

Point Q($\mathbf{0, 1}$) is equidistant from P($\mathbf{5, -3}$) and R($\mathbf{x, 6}$). This means the distance QP is equal to the distance QR. Let's use the distance formula: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$QP = QR$

$\sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2}$

$\sqrt{(5)^2 + (-4)^2} = \sqrt{(x)^2 + (5)^2}$

$\sqrt{25 + 16} = \sqrt{x^2 + 25}$

$\sqrt{41} = \sqrt{x^2 + 25}$ Now, we will square both sides of the equation.

$(\sqrt{41})^2 = (\sqrt{x^2 + 25})^2$

$41 = x^2 + 25$

$x^2 = 41 - 25$

$x^2 = 16$

$x = \pm\sqrt{16}$

$\boxed{x = \pm4}$

Step 2 — Calculate distances QR and PR

We have two possible values for $\mathbf{x}$. Let's calculate the distances for each case.

Case 1: When x = 4

First, let's find the distance QR. R is ($\mathbf{4, 6}$) and Q is ($\mathbf{0, 1}$).

$QR = \sqrt{(4 - 0)^2 + (6 - 1)^2}$

$= \sqrt{(4)^2 + (5)^2}$

$= \sqrt{16 + 25}$

$\boxed{QR = \sqrt{41} \text{ units}}$

Next, let's find the distance PR. P is ($\mathbf{5, -3}$) and R is ($\mathbf{4, 6}$).

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

$= \sqrt{(-1)^2 + (6 + 3)^2}$

$= \sqrt{1^2 + 9^2}$

$= \sqrt{1 + 81}$

$\boxed{PR = \sqrt{82} \text{ units}}$

Case 2: When x = -4

First, let's find the distance QR. R is ($\mathbf{-4, 6}$) and Q is ($\mathbf{0, 1}$).

$QR = \sqrt{(-4 - 0)^2 + (6 - 1)^2}$

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

$= \sqrt{16 + 25}$

$\boxed{QR = \sqrt{41} \text{ units}}$

Next, let's find the distance PR. P is ($\mathbf{5, -3}$) and R is ($\mathbf{-4, 6}$).

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

$= \sqrt{(-9)^2 + (6 + 3)^2}$

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

$= \sqrt{81 + 81}$

$= \sqrt{162}$

$= \sqrt{81 \times 2}$

$= 9\sqrt{2}$

$\boxed{PR = 9\sqrt{2} \text{ units}}$

Answer

(i) The values of $\mathbf{x}$ are $\mathbf{4}$ and $\mathbf{-4}$. (ii) If $\mathbf{x = 4}$, then $\mathbf{QR = \sqrt{41}}$ units and $\mathbf{PR = \sqrt{82}}$ units. (iii) If $\mathbf{x = -4}$, then $\mathbf{QR = \sqrt{41}}$ units and $\mathbf{PR = 9\sqrt{2}}$ units.