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Graph

Sum

Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.

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#### Solution

Let A (-1,1), B (0,5) and C (3,2)

**The equation of line AB is**

y -1 = `(5 -1)/(0+ 1) ("x"+1)`

y = `4"x" + 5`

**The equation of line BC is**

y - 5 = `(2 -5)/(3 -0) ("x" -0)`

y = `-"x"+5`

**The equation of line CA is**

y - 2 = `(1 -2)/(-1 -3) ("x" -3)`

y = `("x")/(4) + (5)/(4)`

Required area = Area of ΔABC

**The equation of line CA is **

y - 2 = `(1 -2)/(-1 -3) ("x" -3)`

y = `("x")/(4) + (5)/(4)`

Required area = Area of ΔABC

= `int_-1^0 (4x + 5) dx + int_0^3 (- x + 5) dx - int_-1^3 ( x/4 + 5/4) dx`

= `[ 2x^2 + 5x ]_-1^0 + [ -x^2/2 + 5x ]_0^3 - [ x^2/8 + 5x/4 ]_-1^3`

= `3 + 21/2 - 39/8 - 9/8`

= `15/2` sq. units

Is there an error in this question or solution?