Question Video: Finding the Equation of a Line given the Equation of a Parallel Line and a Point | Nagwa Question Video: Finding the Equation of a Line given the Equation of a Parallel Line and a Point | Nagwa

# Question Video: Finding the Equation of a Line given the Equation of a Parallel Line and a Point Mathematics • Third Year of Secondary School

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What is the equation of the line parallel to the line (π₯ + 9)/9 = (π¦ + 4)/1 = (π§ β 4)/7 and passing through the point (2, 5, β6)?

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### Video Transcript

What is the equation of the line parallel to the line π₯ plus nine over nine equals π¦ plus four over one equals π§ minus four over seven and passing through the point two, five, negative six?

Alright, so we want to solve for the equation of a line thatβs parallel to the line given, and that also passes through this given point. To get started, we can recognize that the equation of this line is given in a form called Cartesian form. In general, the equation of a line expressed this way looks like this. This line passes through a point with coordinates π₯ one, π¦ one, and π§ one, and itβs parallel to a vector with components π, π, and π.

That last bit is important because recall that we want to solve for the equation of a line parallel to the line given. That means if we look at the denominators of these fractions, we can pick out components of a vector parallel to this line. If we call that vector π, we see it has an π₯-component of nine, a π¦-component of one, and a π§-component of seven. Knowing this, we can now use the fact that this line, whose equation we want to solve for, passes through the point two, five, negative six. And we see that this also is accounted for in the Cartesian form of a line. In this case then, we can say that π₯ one, π¦ one, π§ one equals two, five, negative six.

Weβre now ready to combine this information of the point our line passes through and a vector parallel to it to write it out in Cartesian form. First, weβll say that π₯ minus two divided by nine equals π¦ minus five divided by one. And that equals π§ plus six divided by seven. This is the equation of the line parallel to the given line and passing through the point two, five, negative six.

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