Respuesta :
let's firstly put x + 6y + c = 0, in slope-intercept form, to see what's its slope is.
[tex]\bf x+6y+c=0\implies 6y=-x-c\implies y=\cfrac{-x-c}{6}\implies \\\\\\ y=\stackrel{\downarrow }{-\cfrac{1}{6}}x-\cfrac{c}{6}\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
so, now we know that the line L has a slope of -1/6, therefore then
[tex]\bf L(\stackrel{x_1}{a}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3a}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-4}{3a-a}=\stackrel{\downarrow }{-\cfrac{1}{6}}\implies \cfrac{-1}{2a}=-\cfrac{1}{6} \\\\\\ -6=-2a\implies \cfrac{-6}{-2}=a\implies \boxed{3=a}[/tex]
so, since now we know a = 3, that means the points of L are (3,4) and (9,3), so let's find its equation.
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{3})~\hspace{10em} slope = m\implies -\cfrac{1}{6} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-\cfrac{1}{6}(x-3)\implies y-4=-\cfrac{x}{6}+\cfrac{1}{2}[/tex]
[tex]\bf y=-\cfrac{x}{6}+\cfrac{1}{2}+4\implies y=-\cfrac{x}{6}+\cfrac{9}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6y=-x+27} \\\\\\ x+6y\stackrel{\stackrel{c}{\downarrow }}{-27}=0[/tex]
if you're wondering why we multiplied by the LCD of 6, is just to do away with the denominators.
The equation of a straight line is represented as: [tex]Ax + By = C[/tex].
The values of a and c that makes [tex]x + 6y + c = 0[/tex] a straight line are 3 and -27 respectively.
The points are given as:
[tex](x_1,y_1) = (a,4)[/tex]
[tex](x_2,y_2) = (3a,3)[/tex]
The equation of the line is:
[tex]x + 6y + c = 0[/tex]
Substitute [tex](x_1,y_1) = (a,4)[/tex] for x and y in [tex]x + 6y + c = 0[/tex]
[tex]a + 6 \times 4 + c = 0[/tex]
[tex]a + 24 + c = 0[/tex]
Substitute [tex](x_2,y_2) = (3a,3)[/tex] for x and y in [tex]x + 6y + c = 0[/tex]
[tex]3a + 6 \times 3 + c = 0[/tex]
[tex]3a + 18 + c = 0[/tex]
So, we have:
[tex]a + 24 + c = 0[/tex]
[tex]3a + 18 + c = 0[/tex]
Subtract both equations
[tex]a -3a + 24 - 18 + c - c = 0 - 0[/tex]
[tex]-2a + 6 = 0[/tex]
Solve for a
[tex]-2a = -6[/tex]
Divide by -2
[tex]a= 3[/tex]
Substitute [tex]a= 3[/tex] in [tex]a + 24 + c = 0[/tex]
[tex]3 + 24 + c = 0[/tex]
[tex]27 + c = 0[/tex]
Solve for c
[tex]c = -27[/tex]
Hence, the values of a and c are 3 and -27 respectively.
Read more about linear equations at:
https://brainly.com/question/2263981
