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Question

Question: Find area (in sq units) of the triangle formed by the lines y = 2x, y = 3x and y = 5, A. 25/6 B....

Find area (in sq units) of the triangle formed by the lines y = 2x, y = 3x and y = 5,
A. 25/6
B. 25/12
C. 5/6
D. 17/12
E. 6

Explanation

Solution

we have to first find the intersection points using the equations y = 2x, y = 3x and y = 5. These intersection points will be the coordinates of the triangle. Then, to find the area of a triangle we have to use the determinant formula of the area of the triangle.

Complete step by step answer:
Given: y = 2x
Y = 3x
Y = 5
On solving equation y = 2x and y = 3x, the point of intersection = (0,0)
On solving equation y = 2x and y=5, the point of intersection = (5/2,5)
On solving equation y = 3x and y = 5, the point of intersection = (5/3,5)
By using determinant formula to find area of triangle we get,
Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\\ {{x_2}}&{{y_2}}&1 \\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|
So, according to the above intersection points,
x1=0{x_1} = 0, y1=0{y_1} = 0, x2=52{x_2} = \dfrac{5}{2}, y2=5{y_2} = 5, x3=53{x_3} = \dfrac{5}{3}, y3=5{y_3} = 5
Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1 \\\ {\dfrac{5}{2}}&5&1 \\\ {\dfrac{5}{3}}&5&1 \end{array}} \right|
Area of triangle = 12(256)\dfrac{1}{2}\left( {\dfrac{{25}}{6}} \right)
Area of triangle = 2512\dfrac{{25}}{{12}} sq units

So, the correct answer is “Option B”.

Note: whenever we need to find the point of intersection of the triangle, it is necessary to solve the equation given in question. We can also find the area of the triangle with the method of integration. By drawing the triangle on the graph using the coordinates but determinant method is much easier than the integration method.