Answer :
Gradient of line 1;
y=2x+1
m= 2
Parallel lines have same gradient;
For line 2, gradient = 2
y=mx+c
y=2x+c
Replacing for x and y;
2=2(0)+c
2=c
y=2x+2
y=2x+1
m= 2
Parallel lines have same gradient;
For line 2, gradient = 2
y=mx+c
y=2x+c
Replacing for x and y;
2=2(0)+c
2=c
y=2x+2
Answer: The answer is (b) [tex]y=2x+2.[/tex]
Step-by-step explanation: The equation of a straight line in slope-intercept form is given by
[tex]y=mx+c,[/tex] where, 'm' is the slope and 'c' is the y-intercept of the straight line.
The equation of the given line EF is
[tex]y=2x+1.[/tex]
Here, slope, m=2 and y-intercept, c=1.
Since our new line is parallel to the given line, so the slope of the new line=m=2.
So, let the equation of the new line be
[tex]y=mx+d\\\\\Rightarrow y=2x+d.[/tex]
Now, since the line passes through the point (0,2), so
[tex]2=2\times 0+d\\\Rightarrow d=2.[/tex]
Thus, the equation of the new line parallel to line EF will be
[tex]y=2x+2.[/tex]
The correct option is (b).