# Question Video: Finding the Vector Equation of a Line Given the Cartesian Form Mathematics

The equation of a straight line is given by 𝑦 = 2𝑥 + 5. Which of the following vector equations represents the same line? [A] 𝐫 = ⟨1, 7⟩ + 𝑘⟨2, 4⟩ [B] 𝐫 = ⟨5, 2⟩ + 𝑘⟨2, 4⟩ [C] 𝐫 = ⟨0, −5⟩ + 𝑘⟨2, 2⟩ [D] 𝐫 = ⟨4, 2⟩ + 𝑘⟨2, 5⟩ [E] 𝐫 = ⟨0, 5⟩ + 𝑘⟨1, 3⟩

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### Video Transcript

The equation of a straight line is given by 𝑦 equals two 𝑥 plus five. Which of the following vector equations represents the straight line? Is it (A) 𝐫 equals one, seven plus 𝑘 times two, four? Is it (B) 𝐫 equals five, two plus 𝑘 times two, four? (C) 𝐫 equals zero, negative five plus 𝑘 times two, two. Is it (D) 𝐫 equals four, two plus 𝑘 times two, five? Or (E) 𝐫 equals zero, five plus 𝑘 times one, three.

We’ve been given five vector equations and asked to identify which is equivalent to the Cartesian form 𝑦 equals two 𝑥 plus five. So, let’s begin by reminding ourselves of the general vector equation of a line. The general form is 𝐫 equals 𝐫 sub naught plus 𝑘 times 𝐝. Now, 𝐫 sub naught is the position vector of any point on that line. So, our job here will be to choose the position vector that represents a point that lies on the line 𝑦 equals two 𝑥 plus five.

𝑘 is a scalar constant, and 𝐝 is the direction vector. Now the direction vector is linked very closely to the slope of the line. So, we’ll be able to identify that by comparing it directly to the equation 𝑦 equals two 𝑥 plus five. In fact, let’s begin by identifying the direction vector. To do this, we’ll begin by finding the slope of the line. We know a line of the form 𝑦 equals 𝑚𝑥 plus 𝑐 has a slope of 𝑚. So, in this case, we know the slope of the line is equal to two.

Then, if we know the slope of the line is 𝑚, the direction vector is of the form one, 𝑚. But of course, because this is multiplied by a scalar, we can in fact have any multiple of this. In our case then, the direction vector could be one, two, but it could also be any multiple of this. For instance, we could multiply by two to get the vector two, four. Similarly, we can multiply by negative one to get the vector negative one, negative two. But if we compare these to the examples given in our question, we see that the vector equation in (A) and (B) has a direction vector of two, four.

This means we can instantly disregard options (C), (D), and (E). These direction vectors are not multiples of one, two. And so, they cannot represent the equation of our line. All we now need to identify is whether the position vector one, seven or five, two is suitable for our line. So let’s go back to the equation of our line.

We know 𝑦 equals 𝑚𝑥 plus 𝑐 has a 𝑦-intercept of 𝑐. Well, in this case then, we can say that the 𝑦-intercept must be five. In other words, the line must pass through the point with coordinates zero, five. The position vector of zero, five is the vector that takes us from the origin to this point. So, we could say that the position vector of a point on our line is zero, five. But of course, neither of our position vectors are of this form. So, let’s find another point that lies on this line.

Suppose, for instance, 𝑥 is equal to one. Substituting this into the equation of our line will give us the corresponding 𝑦-coordinate. 𝑦 is two times one plus five, which is equal to seven. This means the point with coordinates one, seven lies on our line, and this has corresponding position vector one, seven. And of course, if we look very carefully, we see that option (A) has this position vector.

But let’s just double-check with option (B). If 𝑥 is equal to five, the corresponding 𝑦-coordinate of the point on our line is two times five plus five, which is 15. This is not the same as the position vector five, two. And so, we can also disregard option (B), thereby confirming that the answer is indeed (A).

The vector equation that represents the line whose Cartesian form is 𝑦 equals two 𝑥 plus five is (A) 𝐫 equals one, seven plus 𝑘 times two, four.