shawqidu19 said:
Homework Statement
The function defined on R by f(x)= 2/3x+2
And cf his graphic representation in an orthonormal mark (o; i; j)
Assuming this is f(x)= (2/3)x+ 2, then its graph is a straight line, passing through (0, 2) and (3, 4).
On the X axis, we consider points A, B and H of respective abscissas 3, 3 and x (x belongs R)
?? A and B are on the x-axis and both have abscissa 3? Then A and B are the same point?
Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x
I'm not certain I am interpreting this right but I
think, since A= (3, 0), C= (3, 4).
Since H= (x,0), M= (x, (2/3)x+ 2).
1°) Make the complete figure ?
As I said above, draw the straight line passing through points (0, 2) and (2, 4).
2°) Calculate the distances AH and MH according to x and by means of the absolute values ?
Okay. A= (2, 0) and H= (x, 0) so AH= |x-2|. M= (x, (2/3)x+ 2). MH= |x- (2/3)x-2|= |(1/3)x- 2|
3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?
The triangle CHA is a right triangle with height AH and Base AC- and AC= |4-2|= 2.
Area is (1/2)base*height
The triangle MOB (O is the origin, (0, 0)?) is
not a right triangle with but we can still calculate its height MH above, and base OH= x. Again area= (1/2)base*height.
Set the two areas equal and solve for x.
Homework Equations
The Attempt at a Solution
You can make me the graphic representation and the rest I shall make him(it) alone please make for me please[/QUOTE]