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The equation of line EF is y = 2x + 1. Write an equation of a line parallel to line EF in slope-intercept form that contains point (0, 2). a. y = -2x - 4. b. y = 2x + 2. c. y = -1/2x - 4. d. y = 1/2x + 2.

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Ckaranja

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

Answer:  The answer is (b) \(y=2x+2.\)


Step-by-step explanation:  The equation of a straight line in slope-intercept form is given by

\(y=mx+c,\) where, 'm' is the slope and 'c' is the y-intercept of the straight line.

The equation of the given line EF is

\(y=2x+1.\)

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

\(y=mx+d\\\\\Rightarrow y=2x+d.\)

Now, since the line passes through the point (0,2), so

\(2=2\times 0+d\\\Rightarrow d=2.\)

Thus, the equation of the new line parallel to line EF will be

\(y=2x+2.\)

The correct option is (b).

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