Area of Triangle in Coordinate Geometry

In this article, we will cover the concept of the area of a triangle. This category falls under the broader category of Coordinate geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept, including one in 2013, one in 2015, one in 2017, two in 2020, five in 2021, two in 2022, and three in 2023.

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This Story also Contains

1. Area of Triangle
2. Area Of Triangle formula using Determinant Method
3. Properties of Triangle
4. Solved Examples Based on the Area of Triangle

Area of Triangle

A triangle is a three-sided polygon that has three edges and three vertices. The area of a triangle is the region enclosed by the three sides of the triangle. The formula for the area of a triangle is $(1 / 2) \times$ base $\times$ altitude.

Area Of Triangle formula using Determinant Method

If vertices of a triangle ABC given as $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ and $\mathrm{C}\left(\mathrm{x}_3, \mathrm{y}_3\right)$, then area of ΔABC is

$\left|\frac{1}{2}\right| \begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}||=\frac{1}{2}\left|\mathrm{x}_1\left(\mathrm{y}_2-\mathrm{y}_3\right)+\mathrm{x}_2\left(\mathrm{y}_3-\mathrm{y}_1\right)+\mathrm{x}_3\left(\mathrm{y}_1-\mathrm{y}_2\right)\right|$

If Points A, B, and C are plotted in the two-dimensional plane and the three points are taken in the anticlockwise sense then the area calculated of the triangle ABC will be positive while if the points are taken in the clockwise sense then the area calculated will be negative. But, if the points are taken arbitrarily, then the area calculated may be positive or negative, the numerical value being the same in all the cases. To avoid this sign problem, we put a modulus sign on the area.

Area of Triangle in Coordinate Geometry

Coordinate Geometry is defined as the study of geometry using coordinate points. The area of a triangle in coordinate geometry can be calculated if the three vertices of the triangle are given in the coordinate plane. The area of a triangle in coordinate geometry is defined as the area or space covered by it in the 2-D coordinate plane.

If vertices of a triangle ABC given as $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$, then area of $\triangle \mathrm{ABC}$ is $1 / 2\left(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right.$

Properties of Triangle

1) If three points A, B, and C are collinear, then the area of the triangle will be zero.

2) The area of the triangle can never be negative.

3) If the coordinates of vertices of triangles are given, then we can also find the area of the triangle by Heron's formula. Calculating side length by distance formula.

Recommended Video Based on the Area of Triangle

Solved Examples Based on the Area of Triangle

Example 1: Let $k$ be an integer such that the triangle with vertices $(k,-3 k),(5, k)$ and $(-k, 2)$ has area 28 sq. units. Then the value of $k$ is

1) 3

2) 1

1) 2

4) None of these

Solution:

Area $=28$ sq units

$
\begin{aligned}
& \text { Area }=\frac{1}{2}|k(k-2)+5(2+3 k)-k(-3 k-k)|=28 \\
& \left|5 k^2+13 k+10\right|=56 \\
& 5 k^2+13 k+10=56 \quad \text { or } \quad 5 k^2+13 k+10=-56 \\
& 5 k^2+13 k-46=0 \quad \text { or } \quad 5 k^2+13 k+66=0
\end{aligned}
$
On solving $k=2$ is the only integral solution
Hence, the answer is the option 3.

Example 2: The points $\left(0, \frac{8}{3}\right),(1,3)$ and $(82,30)$ :

1) Form an obtuse-angled triangle.

2) Form an acute angled triangle.

3) Form a right-angled triangle.

4) Lie on a straight line.

Solution:

Let the points given be $A, B, C$
Area of $\triangle A B C=\frac{1}{2}\left[0(3-30)+1\left(30-\frac{8}{3}\right)+82\left(\frac{8}{3}-3\right)\right]$

$
\begin{aligned}
& =\frac{1}{2}\left[0(27)+\frac{82}{3}-\frac{82}{3}\right] \\
& =0
\end{aligned}
$
Since no triangle is formed.
Hence, points are collinear.
Hence, the answer is the option 4.

Example 3: Let the area of the triangle with vertices $\mathrm{A}(1, \alpha), \mathrm{B}(\alpha, 0)$ and $\mathrm{C}(0, \alpha)$ be 4 sq. units. If the points $(\alpha,-\alpha),(-\alpha, \alpha)$ are collinear, then $B$ is equal to:

Solution:
$
\begin{aligned}
& \text { Area }=\left|\frac{1}{2}[1(-\alpha)+\alpha(\alpha-\alpha)+o(\alpha-0)]\right|=4 \\
& \left|\frac{\alpha}{2}\right|=4 \\
& |\alpha|=8 \\
& \alpha= \pm 8 \\
& \text { also }(\alpha,-\alpha),(-\alpha, \alpha),\left(\alpha^2, \beta\right) \text { are collincar } \\
& \therefore \quad \frac{\alpha+\alpha}{-\alpha-\alpha}=\frac{\beta-\alpha}{\alpha^2+\alpha} \\
& -1=\frac{\beta-\alpha}{64+\alpha} \\
& \beta=-64
\end{aligned}
$
Hence, the answer is the option (3).

Example 4: The area of the triangle whose vertices are $(\mathrm{a} \cos \theta \mathrm{r}, \mathrm{b} \sin \theta \mathrm{r}) ; \mathrm{r}=1,2,3$ is equal to $\mathrm{A}=\mathrm{kab} \sin \left(\frac{\theta_1-\theta_2}{2}\right) \sin \left(\frac{\theta_2-\theta_3}{2}\right) \sin \left(\frac{\theta_3-\theta_1}{2}\right)$ where $\mathrm{k}=$

1) 1

2) 2

3) 3

4) 4

Solution:

$
\begin{aligned}
\text { Area of } \Delta & =\frac{1}{2}\left|\begin{array}{ccc}
\mathrm{a} \cos \theta_1 & \mathrm{a} \sin \theta_1 & 1 \\
\mathrm{a} \cos \theta_2 & \mathrm{a} \sin \theta_2 & 1 \\
\mathrm{a} \cos \theta_3 & \mathrm{a} \sin \theta_3 & 1
\end{array}\right| \\
& =\frac{\mathrm{ab}}{2}\left|\begin{array}{ccc}
\cos \theta_1-\cos \theta_2 & \sin \theta_1-\sin \theta_2 & 0 \\
\cos \theta_2-\cos \theta_3 & \sin \theta_2-\sin \theta_3 & 0 \\
\cos \theta_3 & \sin \theta_3 & 1
\end{array}\right| \\
& =\frac{\mathrm{ab}}{2}\left|\begin{array}{ccc}
2 \sin \frac{\theta_1+\theta_2}{2} \sin \frac{\theta_2-\theta_1}{2} & 2 \sin \frac{\theta_1-\theta_2}{2} \cos \frac{\theta_1+\theta_2}{2} & 0 \\
2 \sin \frac{\theta_2+\theta_3}{2} \sin \frac{\theta_3-\theta_2}{2} & 2 \sin \frac{\theta_2-\theta_3}{2} \cos \frac{\theta_2+\theta_3}{2} & 0 \\
\cos \theta_3 & \sin \theta_3 & 1
\end{array}\right| \\
& =2 \mathrm{ab} \sin \left(\frac{\theta_1-\theta_2}{2}\right) \sin \left(\frac{\theta_2-\theta_3}{2}\right) \sin \left(\frac{\theta_3-\theta_1}{2}\right)
\end{aligned}
$
Hence, the answer is the option (2).