# Calculate the area of the triangle which has as vertices the intercepting points of the graph of f(x) with axis and the origin. f(x)= x-2

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We have the triandle ABC such that:

A(0, 0) B and C are the intercepting points of f(x) and x-axis and y- axis

==> f(x) = x-2 ==> x= 2 ==> B(2,0)

==> f(0) = -2 ==> C ( 0, -2)

Now e have the triangle ABC A(0,0) B(2,0) C(0,-2)

From graphinf we notice that ABC is a right angle triangle where BC is the hypotenuse:

Then, the area A = 1/2 * AB *AC

AB = sqrt(2^2 + 0^2 = 2

AC = sqrt(0^2 + -2^2 = 2

==> A = 1/2 * 2 *2 = 2 square units.

Since a vertex of the triangle is the intercepting point of x axis and y axis, namely the origin, the triangle is right-angled.

If the triangle is right, the area is the semi-product of cathetus.

We'll calculate the x-axis intercepting point of the function f(x).

When the graph is intercepting x axis, y = 0.

y = f(x)

f(x) = 0

x-2 = 0

We'll add 2 both sides:

x = 2

So, the length of the first cathetus of the triangle is 2.

We'll calculate the length of the other cathetus.

When the graph is intercepting y axis, x = 0.

We'll put x = 0 and we'll get:

f(0) = 0-2

y = -2

The other length of the cathetus is -2.

The area is:

A = x*y/2

A = -2*2/2

**A = -2 square units.**

f(x) = x-2 intercepts x axis at x-2 = 0 or x =2: or at A with coordinates (2,0)

f(x) = x-2 intercepts y axis at f(x) = 0-2 = -2 at B with coordinates (0,-2)

So by data the triangle gas vertices O(0,0) ,A(2 ,0) and B(0,-2)

The area of OAB , a right angled triangle with righr angle at the origin O is |(1/2)OA*O| = |(1/2)(2)(-2)| = 2 sq units.