Answer:
A. [tex]y=3x-10[/tex]
C. [tex]y+6=3(x-15)[/tex]
Step-by-step explanation:
Given:
The given line is [tex]6x+18y=5[/tex]
Express this in slope-intercept form [tex]y=mx+b[/tex], where m is the slope and b is the y-intercept.
[tex]6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}[/tex]
Therefore, the slope of the line is [tex]m=-\frac{1}{3}[/tex].
Now, for perpendicular lines, the product of their slopes is equal to -1.
Let us find the slopes of each lines.
Option A:
[tex]y=3x-10[/tex]
On comparing with the slope-intercept form, we get slope as [tex]m_{A}=3[/tex].
Now, [tex]m\times m_{A}=-\frac{1}{3}\times 3=-1[/tex]. So, option A is perpendicular to the given line.
Option B:
For lines of the form [tex]x=a[/tex], where, a is a constant, the slope is undefined. So, option B is incorrect.
Option C:
On comparing with the slope-point form, we get slope as [tex]m_{C}=3[/tex].
Now, [tex]m\times m_{C}=-\frac{1}{3}\times 3=-1[/tex]. So, option C is perpendicular to the given line.
Option D:
[tex]3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}[/tex]
On comparing with the slope-intercept form, we get slope as [tex]m_{D}=-\frac{1}{3}[/tex].
Now, [tex]m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}[/tex]. So, option D is not perpendicular to the given line.
