# What is the area of the triangle formed by the lines 3x + 2y = 8, 2x + y = 1 and the x-axis.

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The area of the triangle formed by the lines 3x + 2y = 8, 2x + y = 1 and the x-axis has to be determined.

First, determine the point of intersection of the lines 3x + 2y = 8 and 2x + y = 1

Substituting y = 1 - 2x in 3x + 2y = 8 gives 3x + 2(1 - 2x) = 8

=> 3x + 2 - 4x = 8

=> -x = 6

=> x = -6

y = 1 + 12 = 13

The distance of the point (-6, 13) from the x-axis is 13.

Now determine the x-intercepts of the two lines.

3x + 2y = 8

The x-intercept is 8/3

2x + y = 1

The x-intercept is 1/2

The distance between 8/3 and 1/2 is 13/6.

Now we have the height of the triangle which is 13 and the base which is 13/6. The area of the triangle is (1/2)*13*(13/6) = 169/12

**The area of the triangle formed by the x-axis, and the lines 3x + 2y = 8 and 2x + y = 1 is 169/12**