Area of Triangle in Coordinate Geometry – Formula, Derivation & Examples

A triangle is a three-sided polygon with three angles and three vertices. It is one of the most studied figures in geometry.  The formula used to calculate the area of a triangle is given by $\frac{1}{2}bh$, where $b$ and $h$ are the base and height of a triangle respectively.

Let’s understand how to find the area of triangle in coordinate geometry if we know the coordinates of its vertices. This formula can also be used to find the area of any polygon if the coordinates of its vertices are known.

Area of Triangle in Coordinate Geometry

For any $\triangle \text{ABC}$ with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$, its area is calculated by using the formula $\text{Area } = \frac{1}{2}|x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2)|$.

area of triangle in coordinate geometry

Derivation of Area of Triangle in Coordinate Geometry

 In coordinate geometry, if we need to find the area of a triangle, we use the coordinates of the three vertices. Let’s consider $\triangle \text{ABC}$ with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$. 

To find the area, let’s draw a few constructions. Draw perpendiculars $\text{AE}$, $\text{CF}$, and $\text{BD}$ from the vertices of the triangle to the horizontal axis. These will form three trapeziums $\text{BAED}$, $\text{ACFE}$, and $\text{BCFD}$.

area of triangle in coordinate geometry

As you can see from the above figure we can express the area of a triangle in terms of the areas of these three trapeziums.

$\text{Area}(\triangle \text{ABC}) = \text{ Area}(\text{Trap. BAED}) + \text{Area}(\text{ Trap.ACFE}) – \text{Area}(\text{ Trap.BCFD})$.

Now, the area of a trapezium in terms of the lengths of the parallel sides (the bases of the trapezium) and the distance between the parallel sides (the height of the trapezium):

Trapezium Area = $\frac{1}{2} \times \text{ Sum of bases } \times \text{ Height}$

Let’s consider anyone trapezium, say $\text{BAED}$. Its bases are $\text{BD}$ and $\text{AE}$, and its height is $\text{DE}$. $\text{BD}$ and $\text{AE}$ can easily be seen to be the $y$ coordinates of $\text{B}$ and $\text{A}$, while $\text{DE}$ is the difference between the $x$ coordinates of $\text{A}$ and $\text{B}$. Similarly, the bases and heights of the other two trapeziums can be easily calculated. 

Thus, we get $\text{Area}(\text{Trap.BAED}) = \frac{1}{2} \times (\text{BD } +\text{ AE}) \times \text{ DE}$.

= $\frac{1}{2} \times (y_2  + y+1) \times (x_1 − x_2)$

$\text{Area}(\text{Trap.ACFE}) = \frac{1}{2} \times (\text{AE} + \text{CF}) \times \text{EF}$

$= \frac{1}{2} \times (y_1  + y_3) \times (x_3  − x_1)$

$\text{Area}(\text{Trap.BCFD}) = \frac{1}{2} \times (\text{BD} + \text{CF}) \times \text{DF}$

$= \frac{1}{2} \times (y_2  + y_3) \times (x_3  − x_2)$

The expression for the area of the triangle in terms of the coordinates of its vertices can thus be given by

$\text{Area}(\triangle \text{ABC}) = \text{Area}(\text{Trap.BAED}) + \text{Area}(\text{Trap.ACFE}) – \text{Area}(\text{Trap.BCFD})$

$= \frac{1}{2} \times [(y_2  + y_1) \times (x_1  − x_2)] + \frac{1}{2} \times [(y_1  + y_3) \times (x_3  − x_1)] – \frac{1}{2} \times [(y_2  + y_3) \times (x_3  − x_2)]$

Now simplifying it we get

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

Taking the $\frac{1}{2}$ common we get

$= \frac{1}{2} (x_1 y_2  − x_2 y_2  + x_1 y_1  − x_2 y_1  − x_3  y_1  − x_1 y_1  + x_3 y_3  − x_1 y_3  − x_3 y_2  + x_2 y_2  – x_3 y_3  + x_2 y_3)$

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

Examples on Area of Triangle in Coordinate Geometry

Example 1: Find area of $\triangle \text{ABC}$ with vertices $\text{A}(−1,2)$, $\text{B}(2,3)$, $\text{C}(4,−3)$.

Area of $\triangle \text{ABC}$ with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ is given by $\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)}$

Therefore area of $\triangle \text{ABC}$ with vertices $\text{A}(−1,2)$, $\text{B}(2,3)$, $\text{C}(4,−3)$ is $\frac{1}{2}{-1 (3  − (-3)) + 2(-3  − 2) + 4(2 − 3)}$

$= \frac{1}{2}{-1 (3  + 3) + 2(-3  − 2) + 4(2 − 3)}$

$= \frac{1}{2}{-1 \times 6 + 2 \times (-5) + 4 \times (-1)}$

$= \frac{1}{2}{-6 – 10 – 4}$

$= \frac{1}{2} \times {-20} = -10$

Since the area of a triangle cannot be negative, therefore the area of $\triangle \text{ABC}$ is $10$ sq units.

Note: Since area cannot be negative, therefore the formula can be written as $|\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)}|$.

Example 2: Find the area of a triangle with the vertices $\text{A}(3,4)$, $\text{B}(4,7)$, and $\text{C}(6,−3)$.

Area of $\triangle \text{ABC}$ with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ is given by $|\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)}|$

Therefore area of triangle with vertices $\text{A}(3,4)$, $\text{B}(4,7)$, and $\text{C}(6,−3)$ is

$|\frac{1}{2}{3 \times (7  − (-3)) + 4 \times ((-3)  − 4) + 6 \times (4  − 7)}|$

$= |\frac{1}{2}{3 \times (7  + 3) + 4 \times (-3  − 4) + 6 \times (4  − 7)}|$

$= |\frac{1}{2}{3 \times 10 + 4 \times (-7) + 6 \times (-3)}|$

$= |\frac{1}{2}{30 – 28 – 18}|$

$= |\frac{1}{2} \times (-16)|$

$= |-8| = 8$ sq units

Checking Collinear Points Using Area of Triangle

The three points $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ are called collinear points, if the three points lie in a straight line. When three points are collinear we cannot draw a triangle connecting these three points. Or we can say that area of a triangle is zero when three points are collinear.

Thus three points $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ are collinear if area of $\triangle \text{ABC}$ is zero, i.e., 

$\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)} = 0$

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

Examples on Checking Collinear Points Using Area of Triangle

Example 1: Examine whether the given points  $\text{A} (3,7)$ and $\text{B} (6,5)$ and $\text{C}(15,-1)$ are collinear.

Three points $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ are collinear, if $x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2) = 0$

Now $3 \times (5  − (-1)) + 6 \times (-1  − 7) + 15 \times (7  − 5) = 3 \times (5  + 1) + 6 \times (-1  − 7) + 15 \times (7  − 5)$

$= 3 \times 6 + 6 \times (-8) + 15 \times 2$

$= 18 – 48 + 30 = 0$

Therefore the points  $\text{A} (3,7)$ and $\text{B} (6,5)$ and $\text{C}(15,-1)$ are collinear.

Example 2: Examine whether the given points  $\text{A} (1,4)$ and $\text{B} (3,-2)$ and $\text{C} (-3,16)$ are collinear.

Three points $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ are collinear, if $x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2) = 0$

$1\times (-2  − 16) + 3 \times (16  − 4) – 3 \times (4  − (-2))  = 1\times (-18) + 3 \times 12 – 3 \times (4  + 2)$

$= -18 + 36 – 3 \times 6$

$= -18 + 36 – 18 = 0$

Therefore the points  $\text{A} (1,4)$ and $\text{B} (3,-2)$ and $\text{C} (-3,16)$ are collinear.

Practice Problems

  1. Find the area of the triangle whose vertices are $(1,−2), (−3,4), (2,3)$.
  2. Find the area of the triangle whose vertices are $(0,0), (10,0), (0,10)$.
  3. Find the value of $p$ when it is given that the area of a triangle is $2$ sq. units and the vertices of the triangle are $\text{A}(1, -2)$, $\text{B}(3, 4)$, and $\text{C}(p,3)$.

FAQs

What is the formula of the area of a triangle in coordinate geometry?

The formula of area of triangle formula in coordinate geometry is $|\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)}|$, where $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are the three vertices of a triangle.

How do you find the area of a triangle with 3 coordinates?

For a $\triangle \text{ABC}$ with coordinates of vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ the area of triangle is given by $|\frac{1}{2}{x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2)}|$.

What is the area of triangle in coordinate geometry collinear?

When three points are collinear then the area of a triangle formed by these points is zero.

How you can check that three points are collinear in coordinate geometry?

The three points $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$ are collinear if $=>x_1 (y_2  − y_3) + x_2(y_3  − y_1) + x_3(y_1  − y_2) = 0$

Conclusion

For any $\triangle \text{ABC}$ with vertices $\text{A}(x_1, y_1)$, $\text{B}(x_2, y_2)$, and $\text{C}(x_3, y_3)$, its area is calculated by using the formula $\text{Area } = \frac{1}{2}|x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2)|$. This formula can also be used to find the area of other polygons and also to check whether the points are collinear.

Recommended Reading

Leave a Comment