Collinearity of Three Points in the Plane

We are given three sets of points and asked to:

1. Plot them on the Cartesian plane.
2. Check whether the three points in each set lie on a single straight line (are collinear).

The sets are:

1. (i) $(1, 3), (-1, -1), (-2, -3)$
2. (ii) $(1, 1), (2, -3), (-1, -2)$
3. (iii) $(0, 0), (2, 2), (5, 5)$

---

Concept: Collinearity of Three Points

Three points $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ are collinear if they lie on the same straight line.

A standard algebraic way to check this is to compare slopes:

• Slope of line $AB$:

$$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_2 \ne x_1)$$

• Slope of line $BC$:

$$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} \quad (x_3 \ne x_2)$$

If $m_{AB} = m_{BC}$, then all three points lie on the same straight line, hence are collinear.

A more general test (that also works when some x-coordinates are equal) uses the area of the triangle formed by the three points. If the area is zero, the points are collinear:

$$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$$

If $\text{Area} = 0$, they are collinear.

In this visualization, we show:

• The three points as neon dots.
• The triangle formed by joining them.
• The area of this triangle, and we numerically test if it is zero (within a small tolerance).
• If the area is near zero, we highlight the segment as a straight line through all three points and mark the case as Collinear.

You can switch between the three given triplets, nudge the coordinates a bit, and see how the shape transitions from a thin triangle to an exact line as the area approaches zero.

---

Step-by-step: Checking Each Set

Let us analyze each set conceptually using the area / slope idea.

(i) Points $(1,3), (-1,-1), (-2,-3)$

Label them:

• $A(1,3)$
• $B(-1,-1)$
• $C(-2,-3)$

Compute slopes:

$$m_{AB} = \frac{-1 - 3}{-1 - 1} = \frac{-4}{-2} = 2$$ $$m_{BC} = \frac{-3 - (-1)}{-2 - (-1)} = \frac{-2}{-1} = 2$$

So $m_{AB} = m_{BC} = 2$. Therefore, the three points are collinear.

(Equivalently, the area of triangle $ABC$ is zero.)

(ii) Points $(1,1), (2,-3), (-1,-2)$

Label them:

• $A(1,1)$
• $B(2,-3)$
• $C(-1,-2)$

Compute slopes:

$$m_{AB} = \frac{-3 - 1}{2 - 1} = \frac{-4}{1} = -4$$ $$m_{BC} = \frac{-2 - (-3)}{-1 - 2} = \frac{1}{-3} = -\tfrac{1}{3}$$

Here $m_{AB} = -4$ and $m_{BC} = -\tfrac{1}{3}$, which are not equal. Therefore, these three points are not collinear.

The area test would give a non-zero value, corresponding visually to a triangle with some thickness.

(iii) Points $(0,0), (2,2), (5,5)$

Label them:

• $A(0,0)$
• $B(2,2)$
• $C(5,5)$

Compute slopes:

$$m_{AB} = \frac{2 - 0}{2 - 0} = 1$$ $$m_{BC} = \frac{5 - 2}{5 - 2} = \frac{3}{3} = 1$$

Thus $m_{AB} = m_{BC} = 1$. Therefore, these three points are collinear.

You might recognize that all points satisfy the equation $y = x$, which is exactly the line passing through these three points.

---

Summary of Answers

• (i) $(1, 3), (–1, –1), (–2, –3)$: Collinear
• (ii) $(1, 1), (2, –3), (–1, –2)$: Not collinear
• (iii) $(0, 0), (2, 2), (5, 5)$: Collinear

---

What the Visualization Shows

• A Cartesian grid centered in the canvas.
• Three points from the currently selected set, shown as bright neon circles.
• The triangle formed by connecting the three points.
• If the computed area is very close to zero, the visualization:
  • Draws a straight neon line passing exactly through the three points.
  • Displays Collinear in neon green-like text.
• If not, it:
  • Shows the triangle with a filled neon face.
  • Displays Not Collinear.

You can:

• Switch between (i), (ii), and (iii).
• Slightly adjust the coordinates of each point to see how a nearly straight configuration becomes exactly collinear.
• Adjust the zoom level to explore the geometry more clearly.

The purpose is to connect the algebraic test (slope / area) with a visual feeling for collinearity: the triangle “collapses” into a line when its area becomes zero.