# Question Video: Finding the Parametric Equations of a Line Passing through a Given Point and Its Direction with Respect to a Given Vector Mathematics

Which of the following are the parametric equations of the line through point 𝐴 (−8, 8) with a direction vector perpendicular to 𝐮 = 〈−6, 7〉? [A] 𝑥 = −8 − 6 𝑘, 𝑦 = 8 + 7𝑘 [B] 𝑥 = −8 + 8𝑘, 𝑦 = −6 + 7𝑘 [C] 𝑥 = −8 + 7𝑘, 𝑦 = 8 − 6𝑘 [D] 𝑥 = −8 + 7𝑘, 𝑦 = 8 + 6𝑘

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### Video Transcript

Which of the following are the parametric equations of the line through point 𝐴 negative eight, eight with a direction vector perpendicular to 𝐮 is equal to the vector negative six, seven? Is it option (A) 𝑥 is equal to negative eight minus six 𝑘, and 𝑦 is equal to eight plus seven 𝑘? Option (B) 𝑥 is equal to negative eight plus eight 𝑘, and 𝑦 is equal to negative six plus seven 𝑘. Is it option (C) 𝑥 is equal to negative eight plus seven 𝑘, and 𝑦 is equal to eight minus six 𝑘? Or is it option (D) 𝑥 is equal to negative eight plus seven 𝑘, and 𝑦 is equal to eight plus six 𝑘?

In this question, we’re asked to determine which of four given options are the parametric equations of a line passing through a given point perpendicular to a given vector. And to do this, let’s start by recalling what we mean by the parametric equations of a line. They’re two equations of the form 𝑥 is equal to 𝑥 sub zero plus 𝑎 times 𝑘 and 𝑦 is equal to 𝑦 sub zero plus 𝑏 multiplied by 𝑘. 𝑘 is called our scalar, and it can take any value. And we know since 𝑘 can be equal to zero, we can substitute this into the equation to note that 𝑥 sub zero, 𝑦 sub zero is a point which lies on the line. And in fact, we can choose any point which lies on the line for the point 𝑥 sub zero, 𝑦 sub zero; this will give us different equations for the line.

Similarly, we can recall that vector 𝑎, 𝑏 is a nonzero vector which is parallel to the line. Therefore, to find the parametric equations of a line, we need a point which lies on the line. And we also need a nonzero vector parallel to the line. And we’re given in the question that the line passes through the point with coordinates negative eight, eight. So, we can choose 𝑥 sub zero to be equal to negative eight and 𝑦 sub zero to be equal to eight. This is one possible choice of a point which lies on the line.

And if we look at our four given options, we can see three of these given options have chosen the same point. The value of 𝑥 sub zero in these equations is negative eight and the value of 𝑦 sub zero is eight. However, in equation (B), we can notice something interesting. The value of 𝑦 sub zero is negative six. It’s possible to use this information to eliminate this option since we know the line is not vertical. And for this line to pass through both 𝐴, which has coordinates negative eight, eight, and this point 𝑥 sub zero, 𝑦 sub zero, which is the point negative eight, negative six, the line would need to be vertical and then it wouldn’t be perpendicular to our vector 𝐮. However, to fully justify this, let’s find a direction vector of this line.

We want to do this by using the fact that our line is perpendicular to the vector 𝐮. And there are many different ways we can use this information to determine a direction vector of the line. For example, for any two-dimensional vector 𝐮, we can determine a vector perpendicular to this vector by switching its components and then multiplying one of the nonzero components by negative one. Switching the components of vector 𝐮 gives us the vector seven, negative six. And then if we switch the sign of the vertical component of this vector, we get the vector seven, six. We can call this vector 𝐝. And for due diligence, let’s check that this vector is in fact perpendicular to 𝐮.

We can do this by recalling two vectors are perpendicular when their dot product is equal to zero. So, we want to find the dot product between vectors 𝐮 and 𝐯 [𝐝]. We need to find the dot product of the vector negative six, seven and the vector seven, six. And we do this by finding the sum of the products of the corresponding components. We get negative six times seven plus seven times six, which we can evaluate is equal to zero. Therefore, we’ve confirmed that these two vectors are perpendicular, and this means that 𝐝 is a direction vector of our line. In fact, any nonzero scalar multiple of 𝐝 will be a direction vector of this line.

So, one possible choice of our direction vector 𝑎, 𝑏 is to set 𝑎 equal to seven and 𝑏 equal to six. And we can see that this is exactly what is written in option (D). The coefficient of 𝑘 in our 𝑥- equation is seven and the coefficient of 𝑘 in the 𝑦-equation is six. And for due diligence, it’s worth noting we can use this fact to eliminate all three of the other options. For example, the vector seven, negative six is not parallel to vector 𝐝, and it’s also not perpendicular to vector 𝐮. We can show this in many different ways. For example, we could show it’s not a scalar multiple of 𝐝. Or we could calculate its dot product with 𝐮, which would be nonzero. And we can also follow the same process to eliminate options (A) and (B).

Therefore, by substituting 𝑥 sub zero is equal to negative eight, 𝑦 sub zero is equal to eight, 𝑎 is equal to seven, and 𝑏 is equal to six into our parametric equations, we were able to show that 𝑥 is equal to negative eight plus seven 𝑘 and 𝑦 is equal to eight plus six 𝑘 is the parametric equations of the line through point 𝐴 is negative eight, eight with a direction vector perpendicular to 𝐮 is the vector negative six, seven. This is option (D).