Question

Pleeeeese help me as fast as you can!!!

Answer 1

Answer:

A. $y=3x-10$

C. $y+6=3(x-15)$

Step-by-step explanation:

Given:

The given line is $6x+18y=5$

Express this in slope-intercept form $y=mx+b$, where m is the slope and b is the y-intercept.

$6x+18y=5\\18y=-6x+5\\y=-\frac{6}{18}x+\frac{5}{18}\\y=-\frac{1}{3}x+\frac{5}{18}$

Therefore, the slope of the line is $m=-\frac{1}{3}$.

Now, for perpendicular lines, the product of their slopes is equal to -1.

Let us find the slopes of each lines.

Option A:

$y=3x-10$

On comparing with the slope-intercept form, we get slope as  $m_{A}=3$.

Now, $m\times m_{A}=-\frac{1}{3}\times 3=-1$. So, option A is perpendicular to the given line.

Option B:

For lines of the form $x=a$, 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  $m_{C}=3$.

Now, $m\times m_{C}=-\frac{1}{3}\times 3=-1$. So, option C is perpendicular to the given line.

Option D:

$3x+9y=8\\9y=-3x+8\\y=-\frac{3}{9}x+\frac{8}{9}\\y=-\frac{1}{3}x+\frac{8}{9}$

On comparing with the slope-intercept form, we get slope as  $m_{D}=-\frac{1}{3}$.

Now, $m\times m_{D}=-\frac{1}{3}\times -\frac{1}{3}=\frac{1}{9}$. So, option D is not perpendicular to the given line.