Respuesta :
so since both lines are vertical to each other, or otherwise namely parallel to each other, then since parallel lines have the same slope, these two also have the same slope.
let's put both in slope-intercept form to see what those equivalent slopes are.
[tex]\bf 3x-(a-3)y-6=0\implies 3x-(a-3)y=6 \\\\\\ -(a-3)y=6-3x\implies y=\cfrac{6-3x}{-(a-3)}\implies y=\cfrac{3x-6}{a-3} \\\\\\ \stackrel{\textit{distributing the denominator}}{y=\stackrel{slope}{\cfrac{3}{a-3}}x-\cfrac{6}{a-3}}\\\\ -------------------------------\\\\ (a+1)x+y-1=0\implies (a+1)x+y=1\implies y=1-(a+1)x \\\\\\ y=\stackrel{slope}{-(a+1)}x+1[/tex]
and since we know the slopes are the same for these parallel lines, then
[tex]\bf \cfrac{3}{a-3}x=-(a+1)x\implies \cfrac{3}{a-3}=-(a+1)\implies 3=-(a+1)(a-3) \\\\\\ 3=-(a^2-2a-3)\implies 3=-a^2+2a+3\implies 0=-a^2+2a \\\\\\ a^2-2a=0\implies a(a-2)=0\implies a= \begin{cases} 0\\ 2 \end{cases}[/tex]
so, if we plug in those values for "a" on either slope expression, we end up with -1 or -3, and those are the two possible slopes with such values.
let's put both in slope-intercept form to see what those equivalent slopes are.
[tex]\bf 3x-(a-3)y-6=0\implies 3x-(a-3)y=6 \\\\\\ -(a-3)y=6-3x\implies y=\cfrac{6-3x}{-(a-3)}\implies y=\cfrac{3x-6}{a-3} \\\\\\ \stackrel{\textit{distributing the denominator}}{y=\stackrel{slope}{\cfrac{3}{a-3}}x-\cfrac{6}{a-3}}\\\\ -------------------------------\\\\ (a+1)x+y-1=0\implies (a+1)x+y=1\implies y=1-(a+1)x \\\\\\ y=\stackrel{slope}{-(a+1)}x+1[/tex]
and since we know the slopes are the same for these parallel lines, then
[tex]\bf \cfrac{3}{a-3}x=-(a+1)x\implies \cfrac{3}{a-3}=-(a+1)\implies 3=-(a+1)(a-3) \\\\\\ 3=-(a^2-2a-3)\implies 3=-a^2+2a+3\implies 0=-a^2+2a \\\\\\ a^2-2a=0\implies a(a-2)=0\implies a= \begin{cases} 0\\ 2 \end{cases}[/tex]
so, if we plug in those values for "a" on either slope expression, we end up with -1 or -3, and those are the two possible slopes with such values.
