Video Transcript
π΄π΅πΆπ· is a quadrilateral, in which π΄ is the point one, two; π΅ is the point one, three; πΆ is the point five, three; and π· is the point five, two. Determine its type using vectors. (A) Parallelogram, (B) trapezoid, (C) rectangle, or option (D) kite.
In this question, weβre given a quadrilateral π΄π΅πΆπ· and weβre given the coordinates of all four of its vertices. We need to determine the type of quadrilateral π΄π΅πΆπ· is, and we need to do this by using vectors. To do this, weβll start by sketching points π΄, π΅, πΆ, and π·. And since weβre given the coordinates of points π΄, π΅, πΆ, and π·, we can even sketch this to scale.
In particular, we note that points π΄ and π΅ share the same π₯-coordinate. However, π΅ is one unit higher than point π΄. Similarly, point πΆ shares the same π¦-coordinate as point π΅. However, its π₯-coordinate is four higher. And finally, point π· is one unit below point πΆ. This gives us a shape which looks like the following. And of course we can determine that all of the angles are right angles because weβre only moving vertically or horizontally.
At this point, it seems weβve sketched a rectangle, which is the correct answer. However, we should check this by using vectors, since weβre told to in the question.
To do this, letβs find the vector from π΄ to π΅. We can find this vector by recalling the vector from π΄ to π΅ is equal to the position vector of π minus the position vector of π. And the position vector of a point has components equal to the coordinates of the point. So we have the vector one, three minus the vector one, two.
And remember, we evaluate the difference between two vectors component-wise. In other words, we need to find the difference in the corresponding components. This gives us the vector one minus one, three minus two. And we can evaluate each component. We get the vector zero, one, which agrees with our diagram. This tells us to move from point π΄ to point π΅, we move one unit upwards.
We can follow the exact same process to find the vector from π· to πΆ. The vector from π· to πΆ will be the position vector of π minus the position vector of π. Thatβs the vector five, three minus the vector five, two. And once again we evaluate the difference between two vectors component-wise. We once again get the vector zero, one. And now we can notice something interesting. The vector from π΄ to π΅ is exactly equal to the vector from π· to πΆ. So, not only are the vectors parallel, we can also note that they have the same length.
Letβs now follow the same process for the vector from π΅ to πΆ and the vector from π΄ to π·. First, the vector from π΅ to πΆ is the position vector of π minus the position vector of π. Thatβs the vector five, three minus the vector one, three. And once again, we evaluate this difference component-wise. We get the vector five minus one, three minus three, which simplifies to give us the vector four, zero. In other words, to move from point π΅ to point πΆ, we move four units right.
And finally we can follow the same process to determine the vector from π΄ to π·. Itβs the vector five, two minus the vector one, two. We evaluate this component-wise and simplify. We get the vector four, zero. So once again we can see something interesting. The vector from π΅ to πΆ is equal to the vector from π΄ to π·. And if the vectors are equal, this means they have the same direction. So the vectors are parallel. They also have the same magnitude, which means the vectors will have the same length.
And it is worth noting we know that side π΅πΆ is four times the length of side π΄π΅. So this is not a square. Therefore, by using vectors, we were able to show the quadrilateral π΄π΅πΆπ· is a rectangle, which is option (C).