[tex]d_1 : ~~~y=3x-5 \\
d_2: ~~~ax+4y=1 \\
\texttt{(Aducem ecuatiile la forma carteziana:)} \\
d_1 : ~~~3x-y-5=0 \\
d_2: ~~~ax+4y-1=0 \\ \\
\texttt{Calculam pantele:} \\
m_1 = - \frac{3}{-1} = 3 \\
m_2 = \frac{-a}{4} \\
a) \\
\texttt{Dreptele sunt paralele daca pantele sunt egale. } \\
m_2 = m_1 \\
\frac{-a}{4} =3 ~~\Longrightarrow~~a = -3 \times 4 = \boxed{-12}
[/tex]
[tex]b) \\
d_1 : ~~~3x-y-5=0 \\
d_2: ~~~-12x+4y-1=0 \\
\text{Pentru a afla intersectia cu axa Ox, il facem pe y=0} \\
3x-5=0 =\ \textgreater \ x= \frac{5}{3} ~~=\ \textgreater \ A1_x(\frac{5}{3}, ~0) \\
-12x-1 = 0 =\ \textgreater \ x= \frac{1}{-12} ~~=\ \textgreater \ A2_x(-\frac{1}{12}, ~0) \\
\text{Pentru a afla intersectia cu axa Oy, il facem pe x=0} \\
-y-5=0 =\ \textgreater \ y=-5 =\ \textgreater \ A1_y(0,~-5) \\
4y-1=0 =\ \textgreater \ y = \frac{1}{4} =\ \textgreater \ A2_y(0,~\frac{1}{4} )
[/tex]
[tex]c) \\
d_1 : ~~~3x-y-5=0 \\
d_2: ~~~ax+4y-1=0 \\
m_1 = - \frac{3}{-1} = 3 \\
m_2 = \frac{-a}{4} \\
\texttt{Dreptele sunt perpendiculare daca: } \\
m2 = -\frac{1}{m1} \\ \\
\frac{-a}{4} = - \frac{1}{3} \\
-a = - \frac{4}{3} \\
a= \frac{4}{3} \\ \\
=\ \textgreater \ d_2: ~~~\frac{4}{3}x+4y-1=0 ~~~=\ \textgreater \ ~~~4x + 12y -3=0
[/tex]
[tex]\text{Pentru a afla punctul lor de intersectie, rezolvam sistemul de ecuatii:} \\
3x-y-5=0 \\
4x + 12y -3=0 \\
--- \\
3x-y=5 ~~~~~=\ \textgreater \ ~~ Substitutis:~~~ \boxed{y = 3x-5 }\\
4x + 12y =3 \\
--- \\
4x+12(3x-5)=3 \\
4x+36x -60 =3 \\
40x = 63 \\
x = \frac{63}{40} \\
\text{Ne intoarcem la substitutie.} \\
y = 3x-5 = 3\times \frac{63}{40} -5 = \frac{3\times 63-200}{40}= \frac{189-200}{40} \\
\text{Punctul de intersectie este:} ~~~
\boxed{B\Big(\frac{63}{40}, ~\frac{-11}{40} \Big)}[/tex]
