B1 the Cartesian Plane Appendix B Review of Graphs Equations and Ineuqlities

In Geometry, a triangle is a 3 – sided polygon which has three edges and 3 vertices. Area of the triangle is a measure out 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 unlike methods used to discover the surface area of a triangle in coordinate geometry.

Methods to Notice the Area of a Triangle

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

Method 1

When the base of operations and altitude of the triangle are given.

Surface area of the triangle, A = bh/2 foursquare 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'due south formula.

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

A =

\(\begin{array}{l} \sqrt{s(due south~-~a)~(south~-~b)~(s~-~c)} \end{array} \)

Where a, b, c are the side lengths of the triangle and southward is the semi perimeter

The value of south is constitute using the formula

south =

\(\brainstorm{array}{fifty} \frac {a~+~b~+~c}{ii} \cease{array} \)

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 detect the surface area of a triangle when the vertices in the coordinate aeroplane is known.

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

From the figure, the expanse of a triangle PQR, lines such as

\(\brainstorm{array}{l} \overleftrightarrow {QA}\end{array} \)

,

\(\begin{assortment}{l} \overleftrightarrow {Lead}\end{assortment} \)

and

\(\begin{array}{l} \overleftrightarrow {RC}\end{array} \)

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 surface area of all the trapeziums.

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

Finding Area of a Trapezium PQAB

We know that the formula to notice the area of a trapezium is

Since Area of a trapezium = (ane/ii) (sum of the parallel sides)×(altitude between them)

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

QA = y₂

PB = y₁

AB = OB – OA = ten₁– ten₂

Expanse of trapezium PQAB = (1/2)(y₁ + y₂)(x_i– x₂ ) —-(2)

Finding Surface area of a Trapezium PBCR

Surface area of trapezium PBCR =(1/ii) (Atomic number 82 + CR) × BC

PB = y₁

CR = y₃

BC = OC – OB =x₃– 10_one

Surface area of trapezium PBCR =(1/2) (y₁+ y₃ )(x₃– x₁ ) —-(iii)

Finding Expanse of a Trapezium QACR

Expanse of trapezium QACR = (1/two)(QA + CR) × Air-conditioning

QA = y_two

CR = y_three

Ac = OC – OA = x₃– ten₂

Area of trapezium QACR =(one/2)(y₂+ y₃ ) (x_three– ten₂ )—-(four)

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

Expanse of ∆PQR = (1/two)[(y_ane+ y₂)(x_i– 10₂ ) + (y₁+ y₃ )(x_iii– x₁) – (y_two+ y₃ ) (ten₃– x_ii )]

A = (1/2) [x₁ (y₂ – y₃ ) + ten_ii (y₃– y_i ) + 10₃(y₁– y₂)]

Special Instance:

If one of the vertices of the triangle is the origin, so 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 surface area of triangle with vertices P(x₁, y_i), Q(x_ii, y₂) and R(ten_iii, y₃) is zero, then (1/2) [x_one (y₂ – y₃ ) + ten₂ (y_three– y₁ ) + x_iii(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 expanse of the ∆ABC whose vertices are A(1, two), B(4, 2) and C(3, 5)?

Solution:

Using the formula,

A =  (1/ii) [x₁ (y_ii – y_iii ) + x₂ (y₃– y_one ) + x_three(y_ane– y₂)]

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

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

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

To know more well-nigh coordinate geometry and areas of polygons in a coordinate aeroplane, log onto www.byjus.com. To lookout interesting videos on the topic, download BYJU'S – The Learning App from Google Play Store.

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Source: https://byjus.com/maths/area-triangle-coordinate-geometry/

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