Can be found by the below Square Matrix. Determinant value of the Matrix yields the surface of the triangle ∆. And it can be denoted by the symbol delta as well - δ

|x1 y1 1|

|x2 y2 1| . 1/2

|x3 y3 1|

Let AL, BM, CN be the ordinates of the points A, B and C respectively.

Now ∆ ABC = trapezium ABML + trap. ALNC - trap BMNC

= 1/2(AL+BM)ML+1/2(AL+CN)LN-1/2(BM+CN)MN

=> ∆ = 1/2 (y1+y2)(x1-x2)+1/2(y1+y3)(x3-x1)-1/2(y2+y3)(x3-x2)

= 1/2[x1y2-x2y1+x3y1-x1y3+x2y3-x3y2]

∴ ∆ = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

i.e., ∆ = 1/2Σx1(y2-y3) numerically.

Note:

|x1 y1 1|

1. ∆=|x2 y2 1| . 1/2

|x3 y3 1|

2. The area obtained by the formula is positive or negative according as the vertices are taken round the triangle anti-clockwise or clockwise

3. If the area of triangle formed by the three points is zero, then the points are

|x1 y1 1|

i.e., if ∆=0, then 1/2.Σx1(y2-y3) = 0 or |x2 y2 1| =0

|x3 y3 1|

is the condition for collinearity of three points.

Source: Polytechnic Mathematics-2, S.Janakiraman, J.J Publications

|x1 y1 1|

|x2 y2 1| . 1/2

|x3 y3 1|

Let AL, BM, CN be the ordinates of the points A, B and C respectively.

Now ∆ ABC = trapezium ABML + trap. ALNC - trap BMNC

= 1/2(AL+BM)ML+1/2(AL+CN)LN-1/2(BM+CN)MN

=> ∆ = 1/2 (y1+y2)(x1-x2)+1/2(y1+y3)(x3-x1)-1/2(y2+y3)(x3-x2)

= 1/2[x1y2-x2y1+x3y1-x1y3+x2y3-x3y2]

∴ ∆ = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

i.e., ∆ = 1/2Σx1(y2-y3) numerically.

Note:

|x1 y1 1|

1. ∆=|x2 y2 1| . 1/2

|x3 y3 1|

2. The area obtained by the formula is positive or negative according as the vertices are taken round the triangle anti-clockwise or clockwise

3. If the area of triangle formed by the three points is zero, then the points are

**collinear**,|x1 y1 1|

i.e., if ∆=0, then 1/2.Σx1(y2-y3) = 0 or |x2 y2 1| =0

|x3 y3 1|

is the condition for collinearity of three points.

Source: Polytechnic Mathematics-2, S.Janakiraman, J.J Publications

## 1 comment:

points are

collinearmeans all the three points are on the same line.Post a Comment