Answer:
(i) The slope of line 'T' is 2.
(ii) The slope of line 'R' is -[tex]\frac{1}{2}[/tex]
Step-by-step explanation:
(i) A line 'T' is parallel to line whose equation is; 3y = 6x + 9
We are to find the slope of this line;
The slopes of two parallel lines is the same.
To put the equation of our second line in cartesian plane format;
3y = 6x + 9
Dividing throughout by 3 we get;
y = 2x + 3
So the slope of our line is 2 and so the slope of the line 'T' is 2.
(ii) Another line 'R' is perpendicular to the line whose equation is 3y = 6x + 9
We are to find the gradient/slope of this line;
The product of any two perpendicular lines is -1.
To put the equation of our second line in cartesian plane format;
3y = 6x + 9
Dividing throughout by 3 we get;
y = 2x + 3
The slope of our second line is 2
Let the slope of line 'R' equal to a
So a × 2 = -1
Hence the slope of our line 'R' = a = [tex]\frac{-1}{2}[/tex] = -[tex]\frac{1}{2}[/tex]
