Area of a triangle – Coordinate geometry – formula

What do you mean by the area of a triangle in a coordinate plane?

In a two-dimensional plane, it is the area enclosed by 3 non-collinear points.

In the given figure, it is the part shaded in blue.

The formula for finding the area

When we know the coordinates of all the three vertices of the triangle, we can use the following formula to find the area.

Area =\frac{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]

Derivation of the above formula

We are going to derive the above formula using the area of trapezium.

Draw perpendicular lines from P, Q and  R (vertices of the triangle) to the x-axis. These perpendicular lines meet the x-axis at A, B and C respectively.

The y coordinates of these points are 0 because they lie on the x-axis.

Now we can see that 

The lines AP, BQ and CR are parallel to each other.
Hence ACQP, BCQR and ABRP are trapeziums having at least one pair of opposite sides parallel.

Area (∆ PQR) + Area (Trapezium ACQP) = Area (Trapezium ABRP) + Area (Trapezium BCQR)

• Area (∆ PQR) = Area (Trapezium ABRP) + Area (Trapezium BCQR) – Area (Trapezium ACQP)

Now we have to calculate the area of these trapeziums using the formula

Area of a trapezium = \frac{1}{2}× (sum of parallel sides) × (Perpendicular distance between them)

Trapezium ABRP

Area = \frac{1}{2}× (sum of parallel sides) × (height)

= \frac{1}{2}× (AP + BR) × (AB)

=\frac{1}{2} \times\left(y_{1}+y_{3}\right) \times\left(x_{3}-x_{1}\right)

Trapezium BCQR

\text { Area }=\frac{1}{2} \times(\text { sum of parallel sides }) \times(\text { height })

=\frac{1}{2} \times(B R+C Q) \times(B C)
=\frac{1}{2} \times\left(y_{2}+y_{3}\right) \times\left(x_{2}-x_{3}\right)

Trapezium ACQP

\text { Area }=\frac{1}{2} \times(\text { sum of parallel sides }) \times(\text { height })

=\frac{1}{2} \times(AP+CQ) \times(AC)
=\frac{1}{2} \times\left(y_{1}+y_{3}\right) \times\left(x_{2}-x_{1}\right)

Area (∆ PQR) = Area (Trapezium ABRP) + Area (Trapezium BCQR) – Area (Trapezium ACQP

=\left[\frac{1}{2} \times\left(y_{1}+y_{3}\right) \times\left(x_{3}-x_{1}\right)\right]+\left[\frac{1}{2} \times\left(y_{2}+y_{3}\right) \times\left(x_{2}-x_{3}\right)\right]-\left[\frac{1}{2} \times\left(y_{1}+y_{2}\right) \times\left(x_{2}-x_{1}\right)\right]

Hence Proved

Approach for solving

1. Try to draw the diagram if not given and analyse it.
2. Then we need to note the values of the given data

• x coordinates of the vertices 
• y coordinates of the vertices

Now just substitute the values in the formula to calculate.

Special Cases

1. If the result of the formula comes negative
   The area of a triangle cannot be negative. Hence, we have to take the numerical value without considering the negative sign.
   The negative value is due to the orientation of vertices i.e. clockwise or anti-clockwise.
2. If the result of the formula comes 0.

This is possible only when the three points are collinear. Hence, they don’t enclose any area.

3. If one of the vertices of the triangle is located at the origin.

Substituting the values x1 = 0 and y1 = 0 in the formula, we get

\text { Area }=\frac{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]

=\frac{1}{2}\left[0\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-0\right)+x_{3}\left(0-y_{2}\right)\right]

=\frac{1}{2}\left[x_{2} y_{3}-x_{3} y_{2}\right]

Examples

• Find the area of a triangle having vertices A (0,3), B (5,6) and C (1,2)?

Solution

\text { Area }= \frac{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]
=\frac{1}{2}[0(6-2)+5(2-3)+1(3-6)]
=(-4)

Since the area can’t be negative
The area of triangle ABC is equal to 4 sq. units

• The area of a triangle having vertices A (0,3), B (5,6) and C (1, m) is 4 sq. units. Find the value of m. (Tricky question)

Solution

In this question since the area is given, we have to consider two cases

Case 1 

4=\frac{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]

\Rightarrow 4=\frac{1}{2}[0(6-m)+5(m-3)+1(3-6)]

\Rightarrow 8=5 m-15-3

\Rightarrow 5 m=26
\Rightarrow m=\frac{26}{5}=5.2

Case 2 

(-4)=\frac{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]
\Rightarrow (-4)=\frac{1}{2}[0(6-m)+5(m-3)+1(3-6)]
\Rightarrow (-8)=5 m-15-3
\Rightarrow  5 m=10
\Rightarrow  m=2

Hence the value of m can be both \frac{26}{5}and 2.

In this question, two triangles are possible as shown in the above figure having the same area.