1. ## Function and Geometry [Eleventh Grade]

The function defined on R by f(x)= (2/3)x+2

And cf his graphic representation in an orthonormal mark (o; i; j)

On the X axis, we consider points A, B and H of respective abscissas 3, -3 and x (x belongs R)

Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x

1°) Make the complete figure ?

2°) Calculate the distances AH and MH according to x and by means of the absolute values ?

3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?

Please Anyone for help me, Can you draw this graphic please please and represents all points A B H M and C

2. Originally Posted by bekridu19
The function defined on R by f(x)= (2/3)x+2

And cf his graphic representation in an orthonormal mark (o; i; j)
On the X axis, we consider points A, B and H of respective abscissas 3, -3 and x (x belongs R)
Or C the point of Cf of the same abscissa as A and M the point of Cf of abscissa x

1°) Make the complete figure ?
2°) Calculate the distances AH and MH according to x and by means of the absolute values ?
3°) To determine for which value (s) of x the areas of triangles CHA and MOB is equal ?

...
to #1: see attachment

to #2:

$\displaystyle |\overline{AH}| = |x-(-3)| = |x+3|$

$\displaystyle |\overline{HM}| = \left|\frac23 x +2\right|$

to #3:

$\displaystyle area_{AHM} = \frac12 \cdot (x+3) \cdot \left(\frac23 x +2 \right)$ . Right triangle, painted red.

$\displaystyle area_{ABM} = \frac12 \cdot 6 \cdot \left(\frac23 x +2 \right)$ Triangle painted blue.

The areas of both triangles are equal if:

$\displaystyle x+3 = 6~\vee~ \frac23 x +2 =0~\implies~\underbrace{x = 3}_{\text{maximum}} ~\vee~ \underbrace{x=-3}_{\text{minimum}}$