Formula for Triangle Area

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.

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