Question Video: Finding the Vector Equation of a Line | Nagwa Question Video: Finding the Vector Equation of a Line | Nagwa

Question Video: Finding the Vector Equation of a Line Mathematics • First Year of Secondary School

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Which of the following is a vector equation for the line that passes through the midpoint of line segment π΄π΅, where π΄ = (5, β2) and π΅ = (3, 8), and point (3, 1)? [A] π« = β¨5, β2β© + πβ¨7, 4β© [B] π« = β¨4, 3β© + πβ¨β1, β2β© [C] π« = β¨β4, β3β© + πβ¨2, β10β© [D] π« = β¨4, 3β© + πβ¨7, 4β© [E] π« = β¨β1, β2β© + πβ¨4, 3β©

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Video Transcript

Which of the following is a vector equation for the line that passes through the midpoint of line segment π΄π΅, where π΄ equals five, negative two and π΅ equals three, eight, and point three, one? Is it (A) vector π« equals five, negative two plus π times seven, four? Is it (B) vector π« equals four, three plus π times negative one, negative two? (C) Vector π« equals negative four, negative three plus π times two, negative 10. (D) π« equals four, three plus π times seven, four. Or is it (E) π« equals negative one, negative two plus π four, three?

Letβs begin by recalling how we represent the vector equation of a straight line. The vector equation of a straight line can be represented in the form π« equals π« sub naught plus π‘ times π. π« sub naught is the position vector of a point on the line. For this reason, π« sub naught can take any number of values. We just need to make sure that the point itself does lie on the line. π‘ is a scalar, a constant scalar. And then π is the direction vector of our line. So, our job is going to be to identify the position vector and the direction vector. We might notice that in this question, each scalar is given as the letter π instead of the letter π‘. And thatβs absolutely fine. Weβll use π throughout the remainder of this question.

So, to find the position vector, letβs identify a single point on the line. We could use the point three, one. But weβre also told that our line passes through the midpoint of line segment π΄π΅. We know that the position vector of point three, one is three, one. And none of the lines in this question have this as their position vector. So thatβs a good indication that weβre going to need to find the midpoint of π΄π΅. To do so, we recall that we can find the midpoint of a pair of coordinates by adding those coordinates and dividing by two. In particular, the midpoint of π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is π₯ sub one plus π₯ sub two over two, π¦ sub one plus π¦ sub two over two.

Weβre looking to find the midpoint of π΄ and π΅, so thatβs the midpoint of five, negative two and three, eight. Thatβs five plus three over two, negative two plus eight over two. Five plus three is eight, and eight divided by two is four. Similarly, negative two plus eight divided by two is three. So, we found the midpoint of line segment π΄π΅. Since the position vector of the point on our line tells us how to get there from the origin, we know the position vector π« sub naught can be given by the vector four, three.

Now that we have the position vector, letβs identify the direction vector. We can find the direction vector by finding the difference between the position vectors of two points on our line. We know that the line passes through three, one and four, three. So, the direction vector could be three, one minus four, three. Itβs worth noting that we can do this in reverse. In other words, we could find four, three minus three, one. When we multiply by the scalar, that scalar can take any value. And so really, the direction vector can be any multiple of this difference.

However, we might notice that if we subtract the vector four, three from three, one, weβll get one of the direction vectors in our question. In fact, subtracting the individual components, we get the vector negative one, negative two. So taking the direction vector to be negative one, negative two and the position vector of a point on our line to be four, three and finally taking the scalar to be π, we get the equation of our line to be four, three plus π times negative one, negative two. Thatβs equivalent to option (B).

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