Earlier we find **Area of Triangle** by using the formula –

Area of Triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

where s = \(a+b+c\over 2\)

and a, b, c are the sides of the triangle.

we have used this heron’s formula to find the area of a triangle when the lengths of its sides are given.

Here, we will learn what is the formula for triangle area in terms of coordinates of its vertices.

## Formula for Triangle Area

The area of triangle, the coordinates of whose vertices are A(\(x_1,y_1\)), B(\(x_2,y_2\)) and C(\(x_3,y_3\)) is given by –

Area of Triangle ABC = \(1\over 2\) |[\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\)]|

Example : Find the area of a triangle whose vertices are A(3, 2), B(11, 8) and C(8, 12).

Solution : Let A = (\(x_1, y_1\)) = (3, 2), B = (\(x_2, y_2\)) = (11, 8) and C = (\(x_3, y_3\)) = (8, 12) be the given points. Then,

Area of Triangle ABC = \(1\over 2\) |[\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\)]|

\(\implies\) A = |[\(3(8-12)+11(12-2)+8(2-8)\)]|

\(\implies\) A = |(-12+110-48)| = 25 sq. units

**Remarks :**

(i) If the area of triangle joining three points is zero, then the points are collinear.

(ii) If altitude of any equilateral triangle is P, then its area = \(P^2\over {\sqrt{3}}\). If ‘a’ be the side of equilateral triangle, then its area = (\(a^2\sqrt{3}\over 4\))

Example : Prove that the area of triangle whose vertices are (t, t-2), (t+2, t+2) and (t+3, t) is independent of t.

Solution : Let A = (\(x_1, y_1\)) = (t, t-2), B = (\(x_2, y_2\)) = (t+2, t+2) and C = (\(x_3, y_3\)) = (t+3, t) be the vertices of given triangle. Then,

Area of Triangle ABC = \(1\over 2\) |[\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\)]|

\(\implies\) A = |[\(t(t+2-t)+(t+2)(t-t+2)+(t+3)(t-2-t-2)\)]|

\(\implies\) A = |(2t+2t+4-4t-12)| = |-4| = 4 sq. units

Clearly, area of triangle ABC is independent of t.