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which equation represents a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1

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Answer:

Equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1 is
\mathbf{y=3x-21}

Explanation:

We need to write equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1

The equation will be in slope-intercept form i.e
y=mx+b where m is slope and b is y-intercept.

Finding Slope

Since the lines are parallel to each other, their slopes will be equal.

Slope of given line: y = 3x + 1 is 3 (Compare it with general equation
y=mx+b we get m = 3)

So, slope of required line is: m=3

Finding y-intercept

Using the point (6,-3) and slope m = 3 we can find y-intercept by using the formula:


y=mx+b\\-3=3(6)+b\\-3=18+b\\b=-3-18\\b=-21

So, we get y-intercept: b= -21

Equation of required line

The equation of required line having slope m=3 and y-intercept b = -21 is:
y=mx+b\\y=3x-21

So, equation of a line that passes through the point ( 6,-3 ) and is parallel to the graph of y = 3x + 1 is
\mathbf{y=3x-21}

User MultiGuy
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