## Coordinate Geometry | Area of A Triangle

The area of a D*ABC* with vertices and is given by area

**Proof :
**

Let and be the vertices of the given D*ABC*.

Draw *AP*, *BM* and *CN* perpendiculars to the x-axis.

Then,

and *MN* =

\ area of D*ABC*

= ar (trap. *BMPA*) + ar (trap.. *APNC*) – ar (trap. *BMNC*)

=

=

=

Since the area is never negative, we have

area

Condition for Collinearity of Three Points

Let the given points be and .

Then, *A*, *B* and *C* are collinear

Þ area of D*ABC* = 0

Þ

Þ

### Question:

#### Find the area of a triangle whose vertices are (1, –1), (–4, 6) and (–3, –5).

### Solution:

The area of the triangle formed by the vertices *A*(1, –1), *B*(–4, 6) and *C*(–3, –5) is given by

=

So, the area of the triangle is 24 square units.

### Question:

#### Find the value of k if the points* A*(2, 3),* B*(4,* k*) and* C*(6, –3) are collinear.

### Solution:

Since the given points are collinear, the area of the triangle will be 0,

Þ

Þ

Þ

Therefore, *k* = 0

### Question:

#### If* A*(–5, 7),* B*(–4, –5),* C*(–1, –6) and* D*(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral* ABCD.*

### Solution:

By joining *B* to *D*, two triangles *ABD* and *BCD *are formed.

Now the area of D*ABD
*

=

= square units

Also, the area of DBCD

=

= square units

So, the area of quadrilateral *ABCD* = 53 + 19 = 72 square units.

### Question:

#### Prove that the points (*a, b + c*), (*b, c + a*) and (*c, a + b*) are collinear.

### Solution:

Let D be the area of the triangle formed by the points (*a, b + c*)*, *(*b, c + a*)* and *(*c, a + b*)

Þ

Þ

Þ D = 0

Since area is zero \ the given points are collinear.

### Question:

#### Find the

value of *k* for which the area formed by the triangle with vertices and is 5 square units.

### Solution:

The vertices of the given D*ABC* are and .

\

and

Area of

=

=

= sq units.

But, area of D*ABC* = 5 sq units [given]

\

Þ

or

Þ

or .

Hence, or