WEBVTT
00:00:02.080 --> 00:00:05.440
Consider the points π΄, π΅, and πΆ in the figure.
00:00:06.400 --> 00:00:13.160
Choose the parametric equations of the line from π΄ to π΅.
00:00:13.800 --> 00:00:17.880
Option (A) π₯ is equal to π and π¦ is equal to one.
00:00:18.560 --> 00:00:22.880
Option (B) π₯ is equal to zero and π¦ is equal to π.
00:00:23.480 --> 00:00:27.960
Option (C) π₯ is equal to π and π¦ is equal to zero.
00:00:28.560 --> 00:00:33.120
Option (D) π₯ is equal to one and π¦ is equal to two.
00:00:33.880 --> 00:00:38.160
Or option (E) π₯ is equal to one, π¦ is equal to one plus π.
00:00:40.160 --> 00:00:47.600
Choose the parametric equations of the line from π΅ to πΆ.
00:00:48.280 --> 00:00:54.120
Option (A) π₯ is equal to π minus three and π¦ is equal to two.
00:00:54.760 --> 00:01:00.160
Option (B) π₯ is equal to π plus one and π¦ is equal to three.
00:01:00.920 --> 00:01:05.320
Option (C) π₯ is equal to two and π¦ is equal to π minus three.
00:01:06.200 --> 00:01:12.480
Option (D) π₯ is equal to three and π¦ is equal to π minus two.
00:01:13.080 --> 00:01:16.600
Or option (E) π₯ is equal to two and π¦ is equal to three.
00:01:17.920 --> 00:01:25.080
Choose the parametric equations of the line from π΄ to πΆ.
00:01:25.720 --> 00:01:33.320
Option (A) π₯ is equal to two π plus one and π¦ is equal to two π plus one.
00:01:33.960 --> 00:01:39.600
Option (B) π₯ is equal to π minus three and π¦ is equal to two.
00:01:40.360 --> 00:01:47.600
Option (C) π₯ is equal to two π minus one and π¦ is equal to three π minus one.
00:01:47.600 --> 00:01:52.880
Option (D) π₯ is equal to two and π¦ is equal to π minus two.
00:01:53.680 --> 00:01:58.120
Or option (E) π₯ is equal to two and π¦ is equal to two.
00:01:58.360 --> 00:02:09.360
In this question, weβre given a coordinate grid and three points on this grid, π΄, π΅, and πΆ.
00:02:10.440 --> 00:02:22.440
We need to use this figure to determine parametric equations for the three lines between the pairs of points given in the figure.
00:02:23.960 --> 00:02:26.680
And thereβs a few different ways of doing this.
00:02:27.440 --> 00:02:33.560
Letβs start by recalling what we mean by the parametric equations of a line.
00:02:34.440 --> 00:02:44.560
Theyβre two equations of the form π₯ is equal to π times π plus π₯ sub zero and π¦ is equal to π times π plus π¦ sub zero.
00:02:45.720 --> 00:02:59.680
And we can recall we can choose any point π₯ sub zero, π¦ sub zero which lies on the line and any nonzero vector π, π which is parallel to the line.
00:03:00.640 --> 00:03:06.440
These will give us equivalent parametric equations for the line.
00:03:07.240 --> 00:03:18.280
For example, in the line between π΄ and π΅, we know that both points π΄ and π΅ lie on the line.
00:03:19.160 --> 00:03:32.640
So, we can choose either of these points to give us our values of π₯ sub zero or π¦ sub zero, or we could choose any point on this line.
00:03:34.040 --> 00:03:43.920
So, instead of choosing this point, letβs instead find a nonzero vector parallel to the line between π΄ and π΅.
00:03:45.240 --> 00:03:48.480
And we can do this in two different ways.
00:03:49.080 --> 00:04:08.600
Either we can notice that the vector from π΄ to π΅ is parallel to our line and we can calculate the vector from π΄ to π΅ by subtracting the position vector of π from the position vector of π or, alternatively, we can directly find the components of this vector from the diagram.
00:04:09.560 --> 00:04:20.840
We see that this line is vertical, so it has no horizontal component, and we can see we move two units up.
00:04:21.760 --> 00:04:29.680
In either case, we show that the vector from π΄ to π΅ is the vector zero, two.
00:04:30.720 --> 00:04:41.200
However, remember we can choose any nonzero vector parallel to the line for our vector ππ.
00:04:42.320 --> 00:04:49.320
So, instead of choosing the vector zero, two, letβs instead choose the vector zero, one.
00:04:50.440 --> 00:04:55.400
Our value of π will be zero, and our value of π will be one.
00:04:56.280 --> 00:05:05.160
And this is exactly the same as noting our line is vertical, so itβs parallel to the unit directional vector π£.
00:05:06.200 --> 00:05:19.200
If we substitute π is equal to zero and π is equal to one into our parametric equations, we see that π₯ will be equal to π₯ sub zero and π¦ will be equal to π plus π¦ sub zero.
00:05:20.400 --> 00:05:27.640
So, we can eliminate any options which donβt have π₯ equal to some constant value.
00:05:28.640 --> 00:05:32.320
These will not be vertical lines.
00:05:33.440 --> 00:05:37.000
And remember here π is a scalar.
00:05:37.720 --> 00:05:41.280
It can take on many values.
00:05:42.240 --> 00:05:46.720
Options (A) and (C) cannot be correct.
00:05:48.000 --> 00:06:02.640
Next, remember our value of π¦ can vary since this is a vertical line, so we canβt have a zero coefficient of π in the π¦ equation.
00:06:03.720 --> 00:06:11.800
Our values of π¦ canβt be constant, so option (D) cannot be correct.
00:06:12.800 --> 00:06:21.120
This leaves us with two remaining options, and we can show that option (B) cannot be the correct answer.
00:06:21.920 --> 00:06:28.920
And to do this, letβs consider the values of π₯ sub zero, π¦ sub zero in equation (B).
00:06:30.000 --> 00:06:36.280
The value of π₯ sub zero is zero, and the value of π¦ sub zero is also zero.
00:06:37.480 --> 00:06:49.840
So, the point with coordinates zero, zero must lie on the parametric equations given in option (B).
00:06:49.840 --> 00:07:01.200
However, we can see that our vertical line does not pass through the origin, so this cannot be the correct answer.
00:07:02.400 --> 00:07:12.560
Instead, letβs choose the point π΄ which has coordinates one, one to be our point π₯ sub zero, π¦ sub zero.
00:07:14.040 --> 00:07:26.600
This means we substitute π΄ is equal to zero, π΅ is equal to one, π₯ sub zero is equal to one, and π¦ sub zero is equal to one into our parametric equations.
00:07:27.920 --> 00:07:35.240
We get π₯ is equal to one and π¦ is equal to one plus π, which we can see is option (E).
00:07:36.360 --> 00:07:47.360
Letβs now clear some space and follow a similar process to determine the parametric equations of the line passing through points π΅ and πΆ.
00:07:48.280 --> 00:07:56.200
This time, we could see on our diagram that the line between π΅ and πΆ is a horizontal line.
00:07:57.160 --> 00:08:11.120
Once again, we can choose either point π΅ or point πΆ or any point on this line to be our point with coordinates π₯ sub zero, π¦ sub zero.
00:08:12.160 --> 00:08:17.600
And we can find a vector parallel to the line in many different ways.
00:08:18.400 --> 00:08:30.880
We could find the vector from π΅ to πΆ, or we can note this is a horizontal line, so itβs parallel to the unit directional vector π’.
00:08:31.840 --> 00:08:36.640
In other words, itβs parallel to the vector one, zero.
00:08:38.040 --> 00:08:44.000
So, weβll choose our value of π΄ equal to one and our value of π΅ equal to zero.
00:08:44.920 --> 00:08:56.720
And since the value of π΅ is zero, we can note in our parametric equations the value of π¦ must be a constant value.
00:08:57.640 --> 00:09:01.640
Options (C) and (D) are incorrect.
00:09:02.840 --> 00:09:10.880
We can also note that option (E) is incorrect since this states that π₯ is a constant value.
00:09:11.720 --> 00:09:17.760
In fact, thereβs only one solution to these parametric equations.
00:09:18.560 --> 00:09:22.720
Itβs the single point with coordinates two, three.
00:09:23.760 --> 00:09:26.440
Itβs not a horizontal line.
00:09:27.520 --> 00:09:30.840
This leaves us with two possible answers.
00:09:31.720 --> 00:09:43.920
And to determine if either of these are possible equations of the line between π΅ and πΆ, we need to determine the coordinates of the point π₯ sub zero, π¦ sub zero.
00:09:45.000 --> 00:09:53.760
And one way of doing this is to note the π¦-coordinate of every single point on our line is three.
00:09:54.720 --> 00:09:59.920
In equation (A), we can see that our value of π¦ is constant.
00:10:00.800 --> 00:10:03.840
However, itβs a constant value of two.
00:10:04.800 --> 00:10:12.080
These are actually the parametric equations of the horizontal line π¦ is equal to two.
00:10:13.280 --> 00:10:16.800
However, we want the horizontal line π¦ is equal to three.
00:10:17.800 --> 00:10:28.520
So, (A) is not the correct answer, and we need π¦ to be the constant value of three, which is given in option (B).
00:10:29.520 --> 00:10:47.680
And although itβs not necessary to conclude this is the correct answer, we can note that the point chosen for π₯ sub zero, π¦ sub zero in these parametric equations is one, three β the point π΅.
00:10:48.840 --> 00:11:14.960
Therefore, substituting π΄ is equal to one, π΅ is equal to zero, π₯ sub zero is equal to one, and π¦ sub zero is equal to three into the parametric equations, we can show that the parametric equations of the line between π΅ and πΆ is π₯ is equal to π plus one and π¦ is equal to three, which is option (B).
00:11:16.200 --> 00:11:22.440
Letβs now do this one final time for the line between π΄ and πΆ.
00:11:23.520 --> 00:11:27.840
Letβs start by finding a vector parallel to this line.
00:11:28.760 --> 00:11:30.920
We can do this from the diagram.
00:11:31.600 --> 00:11:42.960
However, in general, weβll want to use the fact that the vector from π΄ to π is equal to the position vector of π minus the position vector of π.
00:11:43.960 --> 00:11:51.200
And we can recall the components of the position vector of a point are equal to its coordinates.
00:11:52.120 --> 00:11:56.800
So, the position vector of π is the vector three, three.
00:11:57.560 --> 00:12:09.560
And the position vector of π is the vector one, one, since π has coordinates three, three and π has coordinates one, one.
00:12:11.040 --> 00:12:15.160
And now, we evaluate the vector subtraction.
00:12:15.880 --> 00:12:20.800
We do this componentwise.
00:12:21.760 --> 00:12:31.280
We get the vector three minus one, three minus one, which we can evaluate is the vector two, two.
00:12:32.200 --> 00:12:41.240
This is the same as saying for every two units we move across on our line, we move two units upwards.
00:12:42.040 --> 00:12:44.640
We can see this in the diagram.
00:12:45.840 --> 00:12:58.720
We could use the same reasoning we did before to rewrite this as the vector one, one for our parametric equations.
00:12:59.840 --> 00:13:12.480
However, if we look at the five given options, we can see in none of these five options are both coefficients of π in the equation is equal to one.
00:13:13.600 --> 00:13:18.120
In option (A), both of these coefficients are two.
00:13:19.320 --> 00:13:24.920
In option (B), the coefficient of π¦ is zero.
00:13:25.800 --> 00:13:31.560
In option (C), the coefficients of π are two and three.
00:13:32.360 --> 00:13:38.960
In option (D), the coefficient of π in the π₯ equation is zero.
00:13:39.880 --> 00:13:46.280
And in option (E), the coefficients of π in both equations are zero.
00:13:47.600 --> 00:13:55.360
So, the only equation where both coefficients of π are the same is option (A).
00:13:56.360 --> 00:14:08.120
And in fact, this is enough to exclude all four of the other given options since remember the vector ππ must be parallel to the given line.
00:14:09.200 --> 00:14:14.440
And remember this also canβt be the zero vector.
00:14:15.600 --> 00:14:21.800
So, the values of πand π are the same and theyβre both nonzero.
00:14:22.960 --> 00:14:25.760
And this is only true for option (A).
00:14:26.880 --> 00:14:34.160
And we can also, for due diligence, check the value of the point π₯ sub zero, π¦ sub zero.
00:14:35.040 --> 00:14:42.440
Itβs the point with coordinates one, one, which we can see is point π΄.
00:14:43.400 --> 00:15:08.600
Therefore, if we substitute π is equal to two, π is equal to two, π₯ sub zero is equal to one, and π¦ sub zero is equal to one into our parametric equations, we get π₯ is equal to two π plus one and π¦ is equal to two π plus one, which we can see is option (A).