4.3 Q-32

Question Statement

Determine if the points $\mathrm{A}(2,3)$, $\mathrm{B}(-1,1)$, and $\mathrm{C}(4,-5)$ are collinear by calculating the area of the triangle $\triangle ABC$ they form.

If the area is zero, the points are collinear; otherwise, they are not.

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Background and Explanation

To check if points are collinear, we calculate the area of the triangle formed by them using the determinant formula. If the area is zero, it implies that the points lie on the same straight line. The determinant formula for the area is based on the coordinates of the points and evaluates the signed area of the triangle.

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Solution

We use the determinant formula for the area of a triangle given by three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$$\Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Substituting the given points $\mathrm{A}(2,3)$, $\mathrm{B}(-1,1)$, and $\mathrm{C}(4,-5)$:

$$\Delta = \frac{1}{2} \left| 2(1 - (-5)) + (-1)(-5 - 3) + 4(3 - 1) \right|$$

Step 1: Simplify the terms inside the absolute value

• $2(1 - (-5)) = 2(6) = 12$
• $(-1)(-5 - 3) = (-1)(-8) = 8$
• $4(3 - 1) = 4(2) = 8$

Step 2: Add the terms

$$\Delta = \frac{1}{2} \left| 12 + 8 + 8 \right| = \frac{1}{2} \left| 28 \right|$$

Step 3: Multiply by $\frac{1}{2}$

$$\Delta = \frac{1}{2} \times 28 = 14$$

Since the area is $14 \neq 0$, the points are not collinear.

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Key Formulas or Methods Used

• Area of a Triangle:

$$\Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

• Collinearity Condition: Points are collinear if and only if the area of the triangle they form is zero.

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Summary of Steps

1. Use the formula for the area of a triangle with coordinates $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$.
2. Substitute the values of the points $\mathrm{A}(2,3)$, $\mathrm{B}(-1,1)$, and $\mathrm{C}(4,-5)$.
3. Simplify the terms inside the determinant.
4. Calculate the absolute value of the result and divide by 2 to find the area.
5. Check the area:
   • If $\Delta = 0$, the points are collinear.
   • If $\Delta \neq 0$, the points are not collinear.

Here, the area is $14$, so the points $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ are not collinear.