# Area of a Triangle In Coordinate Geometry

In Geometry, a triangle is a 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 the area of a triangle is and different methods used to find the area of a triangle in coordinate geometry.

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## 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 the semi 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 in the coordinate plane is known.

Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), respectively.

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 = y_{2}

PB = y_{1}

AB = OB – OA = x_{1 }– x_{2}

Area of trapezium PQAB = (1/2)(y_{1 }+ y_{2})(x_{1 }– x_{2} ) —-(2)

**Finding Area of a Trapezium PBCR**

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

PB = y_{1}

CR = y_{3}

BC = OC – OB =x_{3 }– x_{1}

Area of trapezium PBCR =(1/2) (y_{1 }+ y_{3} )(x_{3 }– x_{1}) —-(3)

**Finding Area of a Trapezium QACR**

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

QA = y_{2}

CR = y_{3}

AC = OC – OA = x_{3 }– x_{2}

Area of trapezium QACR =(1/2)(y_{2 }+ y_{3} ) (x_{3 }– x_{2} )—-(4)

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

Area of ∆PQR = (1/2)[(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 = (1/2) [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 the origin, then the area of the triangle can be calculated using the below formula.

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

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

A = (ad – 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 (1/2) [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.

### 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) [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3 }– y_{1} ) + x_{3}(y_{1 }– y_{2})]

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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