# Area of a Triangle in Co-ordinate Geometry

Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle ABC.

Then the area of triangle ABC, is

$\large = \frac{1}{2}[ x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2)]$

$\large = \frac{1}{2}\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|$

Notes:

(i) If three points (x1, y1), (x2, y2) and (x3, y3) are collinear, then

$\large \frac{1}{2}\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right| = 0$

(ii) Equation of straight line passing through (x1, y1) and (x2, y2) is given by

$\large \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right| = 0$

(iii) In case of polygon with (x1, y1), (x2, y2), …..(xn, yn) the area is given by

$\large \frac{1}{2} |(x_1 y_2 – y_1 x_2) + (x_2 y_3 – y_2 x_3)+ …..+ (x_{n-1} y_n – y_{n-1} x_n ) + (x_n y_1 – y_n x_1)|$

Example: The vertices of a quadrilateral in order are (−1, 4), (5, 6), (2, 9) and (x , x2). The area of the quadrilateral is 15/2 sq. units, then find the point ( x , x2 )

Solution: Area of quadrilateral = 15/2 (given)

= (1/2) |−26 + 33 + 2x2 − 9x + 4x + x2 | = 15/2

= (1/2) |3x2 − 5x + 7| = ± 15/2

3x2 − 5x + 7 = ± 15

3x2 − 5x − 8 = 0 , 3x2 − 5x + 22 = 0

=> x = 8/3 , x = −1

Hence point is or ( −1 , 1)

But ( −1 , 1) will not form a quadrilateral as per given order of the points.
Hence the required point is (8/3 , 64/9)

Exercise :
(i) Prove that the centroid of the triangle whose vertices are given by A(x1, y1), B(x2, y2) and C(x3, y3) respectively is

$\displaystyle (\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} )$

(ii) Prove that the area of the triangle with vertices at (p − 4 , p + 5), (p + 3 , p − 2) and (p , p) remains constant as p varies and explain the result.

(iii) Show that the lines 3x − 4y + 5 = 0, 7x − 8y + 5 = 0 and 4x + 5y = 45 are concurrent.