# Area Of A Triangle In Coordinate Geometry

Area of a triangle

Triangle is the basic 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.

Area of a triangle is found using different methods.

• 1. When base and altitude of the triangle are given.

Area of the triangle, A = $\frac 12 ~×~b~×~h$

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

• 2. When length of three sides of the triangle are given.

Area of the triangle is calculated using the equation,

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

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

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

Now, what if coordinates of three vertices of the triangle are given?

We can find the length of each sides using distance formula and use Heron’s formula. But this procedure could be tedious especially when lengths of sides are irrational numbers. Therefore, an easy way to find the area has to be found. It is discussed below.

Consider ∆PQR whose coordinates of vertices are $P(x_1,y_1), Q(x_2,y_2)~ and~ R(x_3,y_3).$

Figure 1- Area of triangle with vertices P,Q and R

$\overleftrightarrow {QA}$ , $\overleftrightarrow {PB}$ and $\overleftrightarrow {RC}$ are drawn from Q,P and R respectively perpendicular to x – axis.

PQAB, PBCR and QACR are different trapeziums.

Area of ∆PQR is calculated as,

Area of ∆PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR] —-(1)

Since, Area of a trapezium = $\frac 12$ (sum of the parallel sides)×(distance between them)

Area of trapezium PQAB = $\frac 12$ (QA+PB) × AB

QA = y2

PB = y1

AB = OB – OA = $x_1~-~x_2$

Area of trapezium PQAB = $\frac 12~(y_1~+~y_2 )(x_1~-~x_2 )$ —-(2)

Area of trapezium PBCR = $\frac 12$ (PB + CR) × BC

PB = y1

CR = y3

BC = OC – OB = x3-x1

Area of trapezium PBCR = $12~ (y_1~+~y_3 )~×~(x_3~-~x_1 )$ —-(3)

Area of trapezium QACR = $\frac 12$(QA + CR) × AC

QA= y2

CR = y3

AC = OC – OA = x3-x2

Area of trapezium QACR = $\frac 12~(y_2~+~y_3 )~×~(x_3~-~x_2 )$ —-(4)

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

Area of ∆PQR = $\frac 12[(y_1~+~y_2 )(x_1~-~x_2 )~+~(y_1~+~y_3 )~×~(x_3~-~x_1 )~-~(y_2~+~y_3 )~×~(x_3~-~x_2 )]$

A = $\frac 12 [x_1 (y_2~-~y_3 )~+~x_2 (y_3~-~y_1 )~+~x_3 (y_1~-~y_2)]$

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 = $\frac 12$ [0(b – c) + a(d – 0) + c(0 – b)]

A = $\frac {ac~-~bc}{2}$

• If area of triangle with vertices $P (x_1,y_1), Q(x_2,y_2)$ and $R(x_3,y_3)$ is zero, then
$\frac 12 [x_1 (y_2~-~y_3 )~+~x_2 (y_3~-~y_1 )~+~x_3 (y_1~-~y_2 ) ]$ = 0 and the points $P(x_1,y_1), Q(x_2,y_2)$ and $R(x_3,y_3)$ are collinear.

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

Using the formula,

A = $\frac 12 [x_1 (y_2~-~y_3 )~+x_2 (y_3~-~y_1 )~+~x_3 (y_1~-~y_2)]$

A = $\frac 12$ [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]

A = $\frac 12 [-3 ~+~ 12 ]$ = $\frac 92$ square units.

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#### Practise This Question

The area of the triangle with vertices at (a, b+ c), (b, c + a) and (c, a + b) is