Răspuns :
Trebuie sa gasim functia liniara al carei grafic trece prin punctele A si B
A(1,-2)∈Gf⇒f(1)=-2⇒a+b=-2
B(2,4) ∈Gf⇒f(2)=4 ⇒2a+b=4
................-
-a=-6⇒a=6
6+b=-2⇒b=-8
Deci f(x)=6x-8
M(a,a)∈Gf⇒f(a)=a⇒6a-8=a
5a=8
a=8/5⇒M(8/5 , 8/5)
A(1,-2)∈Gf⇒f(1)=-2⇒a+b=-2
B(2,4) ∈Gf⇒f(2)=4 ⇒2a+b=4
................-
-a=-6⇒a=6
6+b=-2⇒b=-8
Deci f(x)=6x-8
M(a,a)∈Gf⇒f(a)=a⇒6a-8=a
5a=8
a=8/5⇒M(8/5 , 8/5)
Salut,
Ecuaţia dreptei AB este:
[tex]\dfrac{x-x_A}{x_B-x_A}=\dfrac{y-y_A}{y_B-y_A},\;sau\;\dfrac{x-1}{2-1}=\dfrac{y-(-2)}{4-(-2)},\;sau\\\\ 6x-6=y+2,\;sau\;6x-y-8=0.[/tex]
Punctul M aparţine dreptei, dacă [tex]6x_M-y_M-8=0,\;sau\;6\alpha-\alpha-8=0.\;De\;aici\;\alpha=\dfrac{8}5,\;deci\;M\left(\dfrac{8}5,\;\dfrac{8}5\right).[/tex]
Green eyes.