Calvin states that it is not possible to find the equation of a line parrallel to y=2 and passing through the point (5,-1) using the point slope form because the equation of the line does not have an x-term. Explain the error in calvins reasoning. Then find the equation of the parallel line.

1 answer

Calvin's reasoning contains an error because he misunderstands the nature of the line \( y = 2 \). The equation \( y = 2 \) represents a horizontal line where \( y \) is always 2 for any value of \( x \). Therefore, any line parallel to \( y = 2 \) must also be a horizontal line, which means its equation will also have the form \( y = k \), where \( k \) is a constant.

In particular, if we want to find a line that is parallel to \( y = 2 \) and passes through the point \( (5, -1) \), we are looking for a horizontal line through the \( y \)-coordinate of the point \( -1 \).

Thus, the equation of the line parallel to \( y = 2 \) and passing through the point \( (5, -1) \) is simply:

\[ y = -1 \]

This line is horizontal and will never cross the line \( y = 2 \) because it maintains a constant \( y \)-value of \(-1\).

In summary, Calvin's assertion that he cannot find the equation using point-slope form is incorrect, as the concept of parallel lines in this context leads directly to identifying their constant \( y \)-value.