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
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.