Question Video: Using Vectors to Determine the Unknown Components of the Vertices of a Trapezoid Mathematics

Trapezoid 𝐴𝐡𝐢𝐷 has vertices 𝐴(10, 11), 𝐡(π‘˜, 8), 𝐢 (4, βˆ’12), and 𝐷(βˆ’2, 6). Given that 𝐀𝐁 βˆ₯ 𝐂𝐃, find the value of π‘˜.

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Video Transcript

Trapezoid 𝐴𝐡𝐢𝐷 has vertices 𝐴 10, 11; 𝐡 π‘˜, eight; 𝐢 four, negative 12; and 𝐷 negative two, six. Given that the vector from 𝐀 to 𝐁 is parallel to the vector from 𝐂 to 𝐃, find the value of π‘˜.

In this question, we’re given the coordinates of four vertices of a trapezoid and one of the coordinates contains an unknown value of π‘˜. We need to use the four given coordinates and the fact that the vector from 𝐀 to 𝐁 is parallel to the vector from 𝐂 to 𝐃 to find the value of π‘˜.

To answer this question, let’s start by recalling how we calculate the vector between two points. We recall for two points 𝑃 and 𝑄 the vector from 𝐏 to 𝐐 is the position vector of 𝐐 minus the position vector of 𝐏. This can be written 𝐎𝐐 minus 𝐎𝐏. And to find the position vector of a point, we just construct a vector with the components equal to the coordinates of the point. We can use this to find expressions for the vector from 𝐀 to 𝐁 and the vector from 𝐂 to 𝐃. Let’s start with the vector from 𝐀 to 𝐁.

First, this is going to be equal to the position vector of 𝐁 minus the position vector of 𝐀. The position vector of 𝐁 will have components equal to the coordinates of the point 𝐡. It will be the vector π‘˜, eight. Similarly, 𝐴 is the point 10, 11. So the position vector of 𝐀 will be the vector 10, 11. Therefore, the vector from 𝐀 to 𝐁 is the vector π‘˜ eight minus the vector 10, 11. We can then simplify this further. Remember, to subtract two vectors, we need to subtract the corresponding components of the two vectors. Subtracting the first component of each vector, we get π‘˜ minus 10. And subtracting the second component of each vector, we get eight minus 11. So the vector from 𝐀 to 𝐁 is the vector π‘˜ minus 10, 8 minus 11. And finally, we can evaluate the second component, eight minus 11 is equal to negative three. Therefore, the vector from 𝐀 to 𝐁 is the vector π‘˜ minus 10, negative three.

We can do exactly the same thing to find the vector from 𝐂 to 𝐃. This will be equal to the position vector of 𝐃 minus the position vector of 𝐂. The components of the position vector of 𝐃 will be equal to the coordinates of 𝐷. This is going to give us the vector negative two, six. Similarly, the components of the position vector of 𝐂 will be equal to the coordinates of 𝐢. That’s the vector four, negative 12. We need to subtract these two vectors. And remember, to subtract two vectors, we subtract them component-wise.

Subtracting the first component of each vector, we get negative two minus four. And subtracting the second component of each vector, we get six minus negative 12. And if we evaluate the expression for each of our components, we get that the vector from 𝐂 to 𝐃 is the vector negative six, 18. Now that we found expressions for both of these vectors, let’s clear some space and then recall what it means for two vectors to be parallel.

We say that two vectors are parallel if they’re scalar multiples of each other. So to say the vector from 𝐀 to 𝐁 is parallel to the vector from 𝐂 to 𝐃 means that there’s some scalar π‘š such that the vector from 𝐀 to 𝐁 is equal to π‘š times the vector from 𝐂 to 𝐃. We can substitute the expressions we found for the vector from 𝐀 to 𝐁 and the vector from 𝐂 to 𝐃 into this equation to find the value of π‘š. Doing this gives us the equation the vector π‘˜ minus 10, negative three will be equal to π‘š multiplied by the vector negative six, 18.

We can then simplify the right-hand side of this equation by remembering to multiply vector by scalar, we just multiply each of the components of the vector by the scalar. Doing this, we get the vector negative six π‘š, 18π‘š. And we know this must be equal to the vector π‘˜ minus 10, negative three. Now since both of these vectors are equal, their corresponding components must be equal. So we can find the value of π‘š by equating the second component of both of these vectors. We get that negative three must be equal to 18π‘š. We can then solve this equation for π‘š by dividing both sides of the equation through by 18. Doing this, we get that the value of π‘š is negative three divided by 18, which simplifies to give us that π‘š is equal to negative one-sixth.

But we’re not done yet. Remember, the question wants us to find the value of π‘˜. And we can find an equation for the value of π‘˜ by equating the first component of our vectors. We have that π‘˜ minus 10 must be equal to negative six π‘š. But we know the value of π‘š. π‘š is equal to negative one-sixth. So we can substitute this into our equation. Doing this, we get that π‘˜ minus 10 is equal to negative six multiplied by negative one-sixth.

We can evaluate the right-hand side of this equation. Negative six multiplied by negative one-sixth is equal to one. So our equation simplifies to give us that π‘˜ minus 10 is equal to one. And we can solve this by adding 10 to both sides of the equation, which gives us that π‘˜ equals 11, which is our final answer. Therefore, we were able to show if 𝐴𝐡𝐢𝐷 is a trapezoid with vertices 𝐴 10, 11; 𝐡 π‘˜, eight; 𝐢 four, negative 12; and 𝐷 negative two, six and the vector from 𝐀 to 𝐁 is parallel to the vector from 𝐂 to 𝐃, then the value of π‘˜ must be 11.

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