Question

Q2
(a) A line "t" is parallel to 3y = 6x + 9. Find the slope of this line "t".

(b) Another line "r" is perpendicular to the line 3y = 6x + 9. Find the gradient of the line "r"​

Answer 1

Answer:

First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.

Step-by-step explanation:

Answer 2

Given:

Line t:

3y = 6x + 9

To find:

slope=?

Find the gradient of the line "r"​=?

Solution:

The formula of the slope  and gradient of the line:

$\bold{y=mx+c}$

$\bold{ gradient \ of \ the \ line = \frac{(\ change \ in \ y-\ coordinate)}{(\ change \ in \ x-\ coordinate)} }\\\\ OR \\\\\bold{gradient \ of \ the \ line = m}$

In question (a)

line t:

$\to 3y=6x+9\\\\\to 3y= 3(2x+3)\\\\\to y= \frac{3(2x+3)}{3}\\\\\to y= 2x+3$

compare the above value with the slope formula then the slope is = 2

The slope (m)= 2

In question (b)

When the line is perpendicular so, the gradient value is:

$\to \bold{gradient = m} \ when \ perpendicular \\\\\to \bold{gradient = \frac{1}{m}}\\\\\to \bold{gradient = \frac{1}{2}}$