CBSE Tutorials

Coordinate Geometry | Area of A Triangle

The area of a DABC with vertices and is given by area

Proof :

Let and be the vertices of the given DABC.

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

Then,

and MN =

\     area of DABC

= 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 DABC = 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 DABD

=

= 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 DABC are and .

\
and

Area of

=

=

= sq units.

But, area of DABC = 5 sq units     [given]

\    

Þ
or

Þ
or .

Hence,          or