# mathIf function f(x)= - 2x + 5 calculate the area of the triangle between x,y axis and the graph of f(x).

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To evaluate the area of the triangle ABC, we must find out first the x and y intercepts of the graph of the function f(x) = -2x + 5.

To calculate x intercept, we'll have to cancel y = 0.

But y = f(x) => f(x) = 0 <=> -2x + 5 = 0 => -2x = -5 => x = 5/2 => x = 2.5

The intercepting point of the line -2x + 5 and x axis is B(2.5 , 0).

To calculate y intercept, we'll have to put x = 0.

f(0) = -2*0 + 5

f(0) = 5

The intercepting point of the line -2x + 5 and y axis is C(0 , 5).

By definition, x axis is perpendicular to y axis,therefore the triangle 0BC is right angled triangle and the area is the half of the product of cathetus OB and OC.

A = OB*OC/2

We'll calculate the lengths of the cathetus OB and OC.

OB = sqrt[(xB-xO)^2 + (yB-yO)^2]

OB = sqrt [(2.5)^2]

OB = 2.5

OC = sqrt[(xC-xO)^2 + (yC-yO)^2]

OC = sqrt 5^2

OC = 5

A = 5*2.5/2

A = (2.5)^2

**The requested area of the right angled triangle bounded by x and y axis and the graph of f(c) is: A = 6.25 square units.**