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Write the point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 include all of your work in your final answer.

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Answer:

The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0

Solution:

The point slope form of the line that passes through the points
\left(x_(1) y_(1)\right) and parallel to the line with slope “m” is given as


y - y_(1) = m\left(x - x_(1)\right) --- eqn 1

Where “m” is the slope of the line.
x_(1) \text { and } y_(1)are the points that passes through the line.

From question, given that slope “m” = -3

Given that the line passes through the points (6,1).Hence we get
x_(1) = 6 ; y_(1) = 1

By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out.

y – 1 = -3(x – 6)

y – 1 = -3x +18

On rearranging the terms, we get

3x + y -1 – 18 = 0

3x + y – 19 = 0

Hence the point slope form of given line is 3x + y – 19 = 0

User Jiri Trecak
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