What is the area of the triangle formed by these lines?

Concept:Area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3), is given by the expression Area = (1/2) [x1 (y2- y3) + x2 (y3-y1) + x3(y1-y2)]Whose expression can be written in the form of a determinant as: \({\rm{Area}} = {\rm{\;}}\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&1\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&1\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&1 \END{array}} \right|\)Calculation:Given lines are y = 3x, y = 6x and y = 9Let,SIDE AB represent line y = 3x,Side AC represent line y = 9 andSide BC represent line y = 6xSo for coordinates of point A, we will CONSIDER sides AB, AC and then solve respective equation of linesWe have,y = 3x and y = 9⇒ 3x = 9∴ x = 3Therefore A = (3, 9)Similarly, for point B,We have,y = 3x and y = 6x⇒ 6x = 3x∴ x = 0Therefore B = (0, 0)An also, for point C,We have,y = 6x and y = 9⇒ 6x = 9∴ x = 3/2Therefore C = (3/2, 9)As we know that, \({\rm{Area}} = {\rm{\;}}\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&1\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&1\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&1 \end{array}} \right|\)\(\therefore {\rm{Area}} = {\rm{\;}}\frac{1}{2}\left| {\begin{array}{*{20}{c}} 3&9&1\\ 0&0&1\\ {\frac{3}{2}}&9&1 \end{array}} \right|\)Expanding along R2, we get\({\rm{Area\;}} = \left| {\frac{1}{2}{\rm{\;}} \times {\rm{\;}}\left[ {0 + 0 - 1\left( {27 - \frac{{27}}{2}} \right)} \right]} \right| = \frac{{27}}{4}\)Hence area is 27/4 square units.