Question 3

Collinearity: Three points are collinear if they lie on the same straight line. This happens when the sum of distances between consecutive points equals the total distance — i.e., $AB + BC = AC$.

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

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

This formula is derived from the Pythagoras theorem, where the distance is the hypotenuse of a right triangle formed by the horizontal and vertical differences between the two points.

We can check for collinearity using the distance formula.

Step 1 — Calculate AB

Let's find the distance between A($\textbf{1}$, $\textbf{5}$) and B($\textbf{2}$, $\textbf{3}$). We use the distance formula.

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

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

$= \sqrt{1 + 4}$

$\boxed{AB = \sqrt{5}}$

Step 2 — Calculate BC

Now, let's find the distance between B($\textbf{2}$, $\textbf{3}$) and C($\textbf{-2}$, $\textbf{-11}$). We apply the distance formula again.

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

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

$= \sqrt{16 + 196}$

$\boxed{BC = \sqrt{212}}$

Step 3 — Calculate CA

Next, we find the distance between C($\textbf{-2}$, $\textbf{-11}$) and A($\textbf{1}$, $\textbf{5}$). Let's use the distance formula one last time.

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

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

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

$= \sqrt{9 + 256}$

$\boxed{CA = \sqrt{265}}$

Step 4 — Check for collinearity

For points to be collinear, the sum of two distances must equal the third. We check if $\textbf{AB + BC = CA}$.

$\sqrt{5} + \sqrt{212} \approx 2.236 + 14.560$

$= 16.796$

We see that $\textbf{CA} = \sqrt{265} \approx \textbf{16.279}$. Clearly, $\textbf{16.796} \neq \textbf{16.279}$. So, $\textbf{AB + BC} \neq \textbf{CA}$. We also check other combinations. $\textbf{AB + CA} = \sqrt{5} + \sqrt{265} \approx \textbf{2.236} + \textbf{16.279} = \textbf{18.515}$. This is not $\textbf{BC}$. $\textbf{BC + CA} = \sqrt{212} + \sqrt{265} \approx \textbf{14.560} + \textbf{16.279} = \textbf{30.839}$. This is not $\textbf{AB}$. Since no sum of two segment lengths equals the third, the points are not collinear.

Answer

The points $\textbf{(1, 5)}$, $\textbf{(2, 3)}$ and $\textbf{(-2, -11)}$ are not collinear.