Area of a Triangle In Coordinate Geometry

In Geometry, a triangle is the 3 – sided polygon which has 3 edges and 3 vertices. Area of the triangle is a measure of the space covered by the triangle in the two-dimensional plane. In this article, let us discuss what is the area of a triangle, and different methods used to find the area of a triangle in the coordinate geometry.

Methods to Find the Area of a Triangle

Area of a triangle can be found using three different methods. The three different methods are discussed below

Method 1

When the base and altitude of the triangle are given.

Area of the triangle, A = bh/2 square units

where b and h are base and altitude of the triangle respectively.

Method 2

When the length of three sides of the triangle are given, the area of a triangle can be found using the Heron’s formula.

Therefore, the area of the triangle is calculated using the equation,

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

Where a,b, c are the side lengths of the triangle and s is half of the perimeter

The value of s is found using the formula

s = \( \frac {a~+~b~+~c}{2} \)

Method 3

If the vertices of a triangle are given, first we have to find the length of three sides of a triangle. The length can be found using the distance formula

The procedure to find the area of a triangle when the vertices are known in the coordinate plane is given below.

Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x1, y1), (x2, y2), (x3, y3) respectively.

 

Area Of A Triangle in Coordinate Geometry

From the figure, the area of a triangle PQR, lines such as \( \overleftrightarrow {QA}\) , \( \overleftrightarrow {PB}\) and \( \overleftrightarrow {RC}\) are drawn from Q,P and R respectively perpendicular to x – axis.

Now, three different trapeziums are formed such as PQAB, PBCR and QACR in the coordinate plane.

Now, calculate the area of all the trapeziums.

Therefore, the area of ∆PQR is calculated as, Area of ∆PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR] —-(1)

Finding Area of a Trapezium PQAB 

We know that, the formula to find the area of a trapezium is

Since, Area of a trapezium = (1/2) (sum of the parallel sides)×(distance between them)

Area of trapezium PQAB = (1/2)(QA+PB) × AB

QA = y2

PB =y1

AB = OB – OA = x1-x2

Area of trapezium PQAB = (1/2)(y1+y2)(x1-x2 ) —-(2)

Finding Area of a Trapezium PBCR

Area of trapezium PBCR =(1/2) (PB + CR) × BC

PB =y1

CR = y3

BC = OC – OB =x3-x1

Area of trapezium PBCR =(1/2) (y1+ y3 )(x3-x1) —-(3)

Finding Area of a Trapezium QACR

Area of trapezium QACR = (1/2) (QA + CR) × AC

QA=y2

CR = y3

AC = OC – OA = x3-x2

Area of trapezium QACR =(1/2)(y2+ y3 ) (x3-x2 )—-(4)

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

Area of ∆PQR = (1/2)[(y1+y2)(x1-x2 )+(y1+ y3 )(x3-x1)-(y2+ y3 ) (x3-x2 )]

A = (1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]

Special Case:

If one of the vertices of the triangle is origin, then

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

A = (1/2)[0(b – c) + a(d – 0) + c(0 – b)]

A =  (ac-bc)/2 

If area of triangle with vertices P(x1, y1), Q(x2, y2) and R(x3, y3) is zero, then (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.

Area of a Triangle in Coordinate Geometry Example

Example: What is the area of the ∆ABC whose vertices are A(1,2), B(4,2) and C(3,5)?

Solution:

Using the formula,

A =  (1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]

A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]

A = (1/2) [-3 +12]=  9/2 square units.

Therefore, the area of a triangle ABC is  9/2 square units.


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