Question

Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).

Answer 1

The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is  21 square unit.

We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate

Equation of line passing through (0,5) and (2,-2) is

y-5 = [(-2-5)/(2-0)](x-0)

=>y-5 = (-7) /2x

=>y= (-7/2x)+5 -------(eq1)

Equation of line passing through (0,5) and (5,1) is

y-5 = [(1-5) / (5-0)](x-0)

=>y-5 = (-4/5)x

=>y = (-4/5)x+5-------(eq2)

Equation of line passing from (2,-2) and (5,1) is

y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)

=>y+2=(3/3)(x-2)

=>y=x-4--------(eq3)

Now, we use definite integration to find the area between the different equation of line.

So, area enclosed between the equations is given by the

area =$\int\limits^5_2$[(-4/5)x+5 - (-7/2)x + 5)dx  + $\int\limits^5_1$[(-4/5)x+5 -(x-4)]dx

=>area=$\int\limits^5_2$(7/2-4/5)x dx + $\int\limits^5_1$((-9/5)x+9)dx

Using properties of integration,$\int\limits x\, dx=x^{2}/2$

=>area=$\int\limits^5_2$(27/10)x dx + $\int\limits^5_1$(-9/5)x+9)dx

=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)

=>area=(27/20)×(25-4) + (-9/5)×24+9×4

=>area = (27×21)/20 + (-216)/5+ 36

=>area=(567/20) - (216/5) + 36

=>area= [(567-261×4)+(36×20)]/20

=>area=[(567-864)+720]/20

=>area=423/20

=>area=21 square unit.

Hence, area of triangle is 21 square unit.

To know more about area of triangle, visit here:

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