Area Of Triangle In Coordinate Geometry

In Geometry, a triangle is a three-sided polygon that has three edges and three vertices. The area of triangle in coordinate geometry is a measure of the space covered by the triangle in the two-dimensional plane. If the three vertices of the triangle are given in the coordinate plane, the area can be determined. In this article, let us discuss what the area of a triangle is, various methods to find the area of the triangle according to the given parameters and formula of area of triangle in coordinate geometry along with solved examples.

Formula of Area of Triangle in Coordinate Geometry

There are three methods for calculating the area of a triangle. The three approaches are described in detail below.

Method 1

When the triangle's base and altitude are known.

Area of the triangle, A=\[\frac{bh}{2}\]square units

Where b and h are the triangle's base and altitude, respectively.

Method 2

When the length of three sides of the triangle is given, then Heron’s Formula is used to calculate the area of a triangle.

Therefore, the equation used to calculate the triangle's area is given below.

A=\[\sqrt{s(s-a)(s-b)(s-c)}\]

The side lengths of the triangle are a, b, and c, and the semi perimeter is s.

The s value is found by using the formula S=\[\frac{a+b+c}{2}\]

Area of Triangle with Coordinates

Method 3

In this method, we will learn a formula to calculate the area of triangle with coordinates. 

If coordinates points are A(x1,y1),B(x2,y2),C(x3,y3) then the area of triangle using coordinates is $\frac{1}{2}$ [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

Derivation of the Formula is given below:

Assume a triangle PQR with the coordinates P, Q, and R given as (x1,y1),(x2,y2),(x3,y3)

respectively.

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From the figure, the area of a triangle PQR, lines such as  \[\underset{QA}{\leftrightarrow},\underset{PB}{\leftrightarrow}And\underset{RC}{\leftrightarrow}\] are drawn from Q, P and R, respectively perpendicular to the x-axis.

In the coordinate plane, three separate trapeziums, such as PQAB, PBCR, and QACR, are now created.

Calculate the region of each trapezium now.

Therefore, the area of ∆PQR can be calculated as Area of ∆PQR=[\text{Area of trapezium PQAB + Area of trapezium PBCR}] -[\text{Area of trapezium QACR}] —-(1)

Finding Area of a Trapezium PQAB 

The formula for calculating the region of a trapezium is as follows:

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

Area of trapezium PQAB = $\frac{1}{2}$ (QA + PB) × AB

QA = y2

PB = y1

AB = OB – OA = x1-x2

Area of trapezium PQAB = $\frac{1}{2}$ (y1+y2)(x1-x2) —-(2)

Finding Area of a Trapezium PBCR

Area of trapezium PBCR = $\frac{1}{2}$ (PB + CR) × BC

PB = y1

CR =y3 

BC = OC – OB = x3-x1

Area of trapezium PBCR = $\frac{1}{2}$ (y1+y3)(x3-x1) (1/2) —-(3)

Finding Area of a Trapezium QACR

Area of trapezium QACR = $\frac{1}{2}$ (QA + CR) × AC

QA = y2

CR = y3

AC = OC – OA = x3-x2

Area of trapezium QACR = $\frac{1}{2}$ (y2+y3)(x3-x2)-(4)

Substituting (2), (3) and (4) in (1),

Area of ∆PQR = $\frac{1}{2}$ [(y1+y2)(x1-x2)+(y1+y3)(x3-x1)-(y2+y3)(x3-x2)

A=$\frac{1}{2}$ [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

Special Case:

If one of the triangle's vertices is the origin, the triangle's area can be determined using the formula below.

Calculate area of a triangle with vertices are (0,0), P(a, b), and Q(c, d) 

A = $\frac{1}{2}$ [0(b – d) + a(d – 0) + c(0 – b)]

A =  (ad – bc)/2 

Area of Triangle with Vertices

When the area of a triangle with vertices P(x1,y1),Q(x2,y2) and R(x3,y3) is zero, then $\frac{1}{2}$ [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0 and the points P(x1,y1),Q(x2,y2) and R(x3,y3)are collinear. They do not form a triangle since they are in a straight line.

Area of triangle with 3 points

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The formula for calculating area can be expressed using determinants, as shown below.

\[\alpha =\frac{1}{2}\begin{vmatrix}

X_{1} &Y_{1}  &1 \\ 

 X_{2}&Y_{2}  & 1\\ 

X_{3} &Y_{3}  & 1

\end{vmatrix}\]

Where denotes the area of a triangle and A(x1,y1),B(x2,y2) and C(x3,y3)represent the vertices of the triangle.

We know the determinant's value can be either negative or positive. Since we are finding an area and it can never be taken as a negative, therefore we should take the absolute value of the determinant.

We use both the positive and negative values of the determinant if the triangle's area is already known.

Area of triangle with 3 points formula is \[\alpha =\frac{1}{2}\][x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

Another form

The following pictorial representation makes writing the above formula very simple.

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Area of ABC=12{(x1.y2+x2y3+x3y1)-(x2y1+x3y2+x1y3) sq. units.

Note: As we know a triangle's area can never be negative. So we must take the absolute value, in case the area happens to be negative.

Area of a Triangle in Coordinate Geometry Example

1. Calculate the area of the ∆ABC whose vertices are A(1, 2), B(4, 2) and C(3, 5)?

Sol: Using the formula,

12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

A=12[1(2-5)+4(5-2)+3(2-2)]

A=12[-3+12]

A=92 square units.

Hence, the area of a triangle ABC is  92 square units.

2. Find the area of the triangle using vertices A(3,4), B(4,7) and C(3,5).

Sol: Here x1=3,y1=4,x2=4,y2=7,x3=3 and y3=5

We know the formula of area of a triangle is given by

12[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

=12[3(7-3)+4((-3)-(-4))+6(4-(-7))]

=1230+4-18

=8 sq. units

Hence, the area of a triangle is 8 unit square

Conclusion:

We have discussed the area of triangle formula in coordinate geometry. When three points are collinear, the triangle's area is zero. The area of a triangle cannot be negative. When we get the answer in negative terms, we should consider the numerical value of the area, without the negative sign as the area is always positive.