Area of a Triangle and Collinearity in 2D

Formula for Area of a Triangle using Vertex Coordinates

In coordinate geometry, finding the area of a triangle is a common task. If the coordinates of the three vertices of a triangle are known, we can calculate its area directly using a specific formula, without needing to calculate side lengths or angles beforehand. This method is particularly useful when dealing with complex figures or verifying properties analytically.

Let the vertices of a triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

Derivation using Trapezoids

One common method to derive the area formula is by utilizing the concept of signed areas of trapezoids formed by dropping perpendiculars from the vertices to one of the coordinate axes (usually the x-axis). The area of the triangle can be expressed as the sum or difference of the areas of these trapezoids.

Let's assume, for simplicity in the diagram, that the x-coordinates of the vertices are ordered such that $x_1 < x_2 < x_3$. Draw perpendiculars from the vertices A, B, and C to the x-axis, meeting the x-axis at points L, M, and N respectively. The coordinates of these points on the x-axis will be $L(x_1, 0)$, $M(x_2, 0)$, and $N(x_3, 0)$.

Observe the figure. The area of triangle ABC can be seen as the area of the trapezoid ALMB plus the area of the trapezoid BMNC, minus the area of the trapezoid ALNC.

Area($\triangle ABC$) = Area(Trapezoid ALMB) + Area(Trapezoid BMNC) - Area(Trapezoid ALNC)

Recall the formula for the area of a trapezoid: $\frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Distance between parallel sides})$.

• Trapezoid ALMB: The parallel sides are AL (length $|y_1|$) and BM (length $|y_2|$). The height is LM (length $|x_2 - x_1|$). Assuming $y_1, y_2 > 0$ and $x_2 > x_1$:
  Area(ALMB) $= \frac{1}{2} (AL + BM) \times LM = \frac{1}{2} (y_1 + y_2)(x_2 - x_1)$.
• Trapezoid BMNC: The parallel sides are BM (length $|y_2|$) and CN (length $|y_3|$). The height is MN (length $|x_3 - x_2|$). Assuming $y_2, y_3 > 0$ and $x_3 > x_2$:
  Area(BMNC) $= \frac{1}{2} (BM + CN) \times MN = \frac{1}{2} (y_2 + y_3)(x_3 - x_2)$.
• Trapezoid ALNC: The parallel sides are AL (length $|y_1|$) and CN (length $|y_3|$). The height is LN (length $|x_3 - x_1|$). Assuming $y_1, y_3 > 0$ and $x_3 > x_1$:
  Area(ALNC) $= \frac{1}{2} (AL + CN) \times LN = \frac{1}{2} (y_1 + y_3)(x_3 - x_1)$.

Substituting these into the expression for Area($\triangle ABC$). It's important to use the directed distances $(x_2-x_1)$, $(x_3-x_2)$, $(x_3-x_1)$ and signed y-coordinates in a more rigorous derivation using signed areas of trapezoids. The formula naturally emerges without absolute values initially, and then the final absolute value is taken. Let's use the coordinates directly:

Area($\triangle ABC$) $= \frac{1}{2} (y_1 + y_2)(x_2 - x_1) + \frac{1}{2} (y_2 + y_3)(x_3 - x_2) - \frac{1}{2} (y_1 + y_3)(x_3 - x_1)$

Factor out $\frac{1}{2}$:

Area($\triangle ABC$) $= \frac{1}{2} [ (y_1 + y_2)(x_2 - x_1) + (y_2 + y_3)(x_3 - x_2) - (y_1 + y_3)(x_3 - x_1) ]$

Expand the products:

Area($\triangle ABC$) $= \frac{1}{2} [ (y_1x_2 - y_1x_1 + y_2x_2 - y_2x_1) + (y_2x_3 - y_2x_2 + y_3x_3 - y_3x_2) - (y_1x_3 - y_1x_1 + y_3x_3 - y_3x_1) ]$

Remove the inner parentheses, distributing the negative sign for the last term:

Area($\triangle ABC$) $= \frac{1}{2} [ y_1x_2 - y_1x_1 + y_2x_2 - y_2x_1 + y_2x_3 - y_2x_2 + y_3x_3 - y_3x_2 - y_1x_3 + y_1x_1 - y_3x_3 + y_3x_1 ]$

Cancel out the terms that appear with opposite signs ($y_1x_1$, $y_2x_2$, $y_3x_3$):

Area($\triangle ABC$) $= \frac{1}{2} [ y_1x_2 - y_2x_1 + y_2x_3 - y_3x_2 - y_1x_3 + y_3x_1 ]$

Rearrange the terms by grouping by x-coordinates or y-coordinates:

Area($\triangle ABC$) $= \frac{1}{2} [ x_1y_2 - y_1x_2 + x_2y_3 - y_2x_3 + x_3y_1 - y_3x_1 ]$

This can be written more symmetrically by factoring out $x_i$ terms:

Since area must be a non-negative quantity, we take the absolute value of the expression obtained in equation (i):

Formula for Area of a Triangle:

The area of the triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is given by:

$\mathbf{\text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |}$

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Alternative Forms: Determinant Form and Shoelace Formula

The area formula can be conveniently expressed and remembered using a determinant or the Shoelace formula, which is derived from the determinant form.

Determinant Form:

The area of the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ can be represented as half the absolute value of the determinant of a $3 \times 3$ matrix:

$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

Expanding this determinant along the first row gives:

Area $= \frac{1}{2} | x_1(y_2 \cdot 1 - y_3 \cdot 1) - y_1(x_2 \cdot 1 - x_3 \cdot 1) + 1(x_2 y_3 - x_3 y_2) |$

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

Distribute the $-y_1$ term:

Area $= \frac{1}{2} | x_1(y_2 - y_3) - y_1x_2 + y_1x_3 + x_2 y_3 - x_3 y_2 |$

Rearranging terms to match the standard formula:

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

This confirms the equivalence of the determinant form and the standard formula.

Shoelace Formula (or Surveyor's Formula):

This provides a visual and systematic way to compute the area from the coordinates. It is particularly useful for polygons with more than three vertices as well.

1. List the coordinates of the vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ in a column, repeating the first vertex's coordinates at the end.

$\begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \\ x_3 & y_3 \\ x_1 & y_1 \end{pmatrix}$

2. Multiply diagonally downwards and sum these products:

$S_1 = x_1y_2 + x_2y_3 + x_3y_1$

3. Multiply diagonally upwards and sum these products:

$S_2 = y_1x_2 + y_2x_3 + y_3x_1$

4. The area is half the absolute difference between the two sums $S_1$ and $S_2$.

$\mathbf{\text{Area} = \frac{1}{2} | S_1 - S_2 | = \frac{1}{2} | (x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1) |}$

Expanding the Shoelace formula: $\frac{1}{2} | x_1y_2 + x_2y_3 + x_3y_1 - y_1x_2 - y_2x_3 - y_3x_1 |$.

Rearranging terms: $\frac{1}{2} | (x_1y_2 - y_1x_2) + (x_2y_3 - y_2x_3) + (x_3y_1 - y_3x_1) |$. This is the same as the standard formula, just grouped differently inside the absolute value.

The absolute value is essential because the order in which the vertices are listed (clockwise or counter-clockwise) affects the sign of the result before taking the absolute value. Listing them counter-clockwise usually gives a positive result, while a clockwise order gives a negative result, but the absolute value makes the area positive in both cases.

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Answer:

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Formula Summary (For Competitive Exams)

For a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

Standard Form:

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

Determinant Form:

Area $= \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

Shoelace Form:

List coordinates $\begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \\ x_3 & y_3 \\ x_1 & y_1 \end{pmatrix}$. Calculate $S_1 = x_1y_2 + x_2y_3 + x_3y_1$ (downward products sum) and $S_2 = y_1x_2 + y_2x_3 + y_3x_1$ (upward products sum).

Area $= \frac{1}{2} | S_1 - S_2 | = \frac{1}{2} | (x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1) |$

Key Point:

Always remember to take the absolute value of the result, as area is always a non-negative quantity. If the calculated value before taking the absolute value is zero, it indicates that the three points are collinear.

Condition for Collinearity of Three Points using Area Formula

In coordinate geometry, a set of three or more points are defined as collinear if they all lie on the same single straight line. If three points are not collinear, they will form a triangle.

Concept using Area of a Triangle

Consider three distinct points in the Cartesian plane: $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

If these three points A, B, and C are collinear, it means they lie on the same straight line. In this scenario, they cannot form a closed geometric figure with a measurable area, such as a triangle. Visually, the "triangle" formed by collinear points degenerates into a line segment (or a point if all points are the same).

Therefore, if points A, B, and C are collinear, the area of the triangle formed by these three points must be zero.

Conversely, if the area of the triangle formed by three points is calculated to be zero, it implies that these points do not form a triangle and must lie on the same straight line.

Thus, the condition for three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ to be collinear is that the area of $\triangle ABC$ is equal to zero.

Derivation of the Collinearity Condition using Area Formula

We use the formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Area($\triangle ABC$) $= \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

For the points to be collinear, the Area($\triangle ABC$) must be 0.

An absolute value is zero if and only if the expression inside the absolute value is zero. Also, multiplying by 2 doesn't change whether the expression is zero.

So, from equation (i), the condition for collinearity is:

This is the fundamental algebraic condition for the collinearity of three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$.

Determinant Form of the Condition:

Recall the determinant form of the area of a triangle:

Area = $\frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

Setting the area to zero gives:

$\frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| = 0$

This simplifies to the condition that the determinant must be zero:

$\mathbf{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0}$

Expanding this determinant gives the same algebraic condition shown in equation (ii).

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Alternative Method: Using Slopes

Another common method to check for the collinearity of three points relies on the concept of slopes.

If three distinct points A, B, and C are collinear, they all lie on the same straight line. A unique property of a straight line is that its slope is constant throughout. Therefore, the slope of the line segment joining any two of these points must be equal to the slope of the line segment joining any other pair of these points (provided the segments are not vertical).

Consider points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

The slope of the line segment AB is $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$ (if $x_1 \neq x_2$).

The slope of the line segment BC is $m_{BC} = \frac{y_3 - y_2}{x_3 - x_2}$ (if $x_2 \neq x_3$).

The slope of the line segment AC is $m_{AC} = \frac{y_3 - y_1}{x_3 - x_1}$ (if $x_1 \neq x_3$).

The condition for collinearity using slopes is that the slope of AB must be equal to the slope of BC (or $m_{AB} = m_{AC}$ or $m_{BC} = m_{AC}$). We choose $m_{AB} = m_{BC}$:

provided that $x_1 \neq x_2$ and $x_2 \neq x_3$.

Let's cross-multiply equation (iii) to see its relation to the area condition:

$(y_2 - y_1)(x_3 - x_2) = (y_3 - y_2)(x_2 - x_1)$

Expand both sides:

$y_2x_3 - y_2x_2 - y_1x_3 + y_1x_2 = y_3x_2 - y_3x_1 - y_2x_2 + y_2x_1$

Move all terms to one side (e.g., left side):

$y_2x_3 - y_2x_2 - y_1x_3 + y_1x_2 - y_3x_2 + y_3x_1 + y_2x_2 - y_2x_1 = 0$

Cancel out terms ($y_2x_2$ and $-y_2x_2$):

$y_2x_3 - y_1x_3 + y_1x_2 - y_3x_2 + y_3x_1 - y_2x_1 = 0$

Rearrange the terms, grouping by $x_1, x_2, x_3$:

$x_1(y_3 - y_2) + x_2(y_1 - y_3) + x_3(y_2 - y_1) = 0$

This is the same condition as the area condition, just with the terms within the parentheses negated (which is equivalent to multiplying the whole equation by -1, and $0 \times -1 = 0$). So, the slope method is algebraically equivalent to the area method when slopes are defined.

Limitation of the Slope Method:

The slope method requires the denominators $(x_2 - x_1)$ and $(x_3 - x_2)$ to be non-zero. This means it cannot be directly applied if the line is vertical (where $x_1 = x_2 = x_3$). However, if $x_1 = x_2 = x_3$, the points are collinear on a vertical line. If, for example, $x_1 = x_2$ but $x_1 \neq x_3$, the slope of AB is undefined. In this case, A and B lie on a vertical line. For C to be collinear with A and B, C must also lie on this same vertical line, meaning $x_3$ must be equal to $x_1$ (and $x_2$). The area formula $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$ works universally for all orientations of the line.

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Answer:

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