WEBVTT
00:00:02.640 --> 00:00:11.710
Determine the area of the triangle bounded by the π₯-axis, the π¦-axis, and the straight line two π₯ plus seven π¦ plus 28 equals zero.
00:00:12.300 --> 00:00:17.690
Letβs begin by drawing the line two π₯ plus seven π¦ plus 28 equals zero.
00:00:18.020 --> 00:00:23.650
We can create a table of values to find the coordinates where this line will cross both axes.
00:00:24.150 --> 00:00:29.120
It may be helpful to put this equation into the form π¦ equals ππ₯ plus π.
00:00:29.510 --> 00:00:42.730
Rearranging to have just seven π¦ on one side means that we must subtract two π₯ and subtract 28 from both sides of the equation, which gives us seven π¦ equals negative two π₯ minus 28.
00:00:43.260 --> 00:00:49.780
Dividing through by seven then, we have π¦ equals negative two-sevenths π₯ minus 28 over seven.
00:00:50.010 --> 00:00:53.580
And we can notice that this final fraction will simplify to four.
00:00:53.780 --> 00:00:59.940
This form of the equation will allow us to more easily find the points where the line crosses the π₯- and π¦-axes.
00:01:00.500 --> 00:01:07.640
To determine the point where a line crosses the π¦-axis, this could be determined when the π₯-value of the coordinate is equal to zero.
00:01:08.000 --> 00:01:22.500
So substituting π₯ equals zero into our equation π¦ equals negative two-sevenths π₯ minus four gives us π¦ equals negative two-sevenths times zero minus four, which simplifies to give us π¦ equals negative four.
00:01:22.810 --> 00:01:29.680
Recall that, in the general form of an equation π¦ equals ππ₯ plus π, the π-value relates to the π¦-intercept.
00:01:29.740 --> 00:01:32.320
Here, this will be equal to negative four.
00:01:32.920 --> 00:01:39.520
The coordinate where the line crosses the π₯-axis can be determined by finding the value when π¦ is equal to zero.
00:01:39.850 --> 00:01:47.580
So substituting π¦ equals zero into our equation, we have zero equals negative two-sevenths π₯ minus four.
00:01:47.900 --> 00:01:56.260
Rearranging to get π₯ by itself, we can begin by adding four to both sides, which gives four equals negative two-sevenths π₯.
00:01:56.830 --> 00:02:02.270
We can then multiply through by seven to give us 28 equals negative two π₯.
00:02:02.750 --> 00:02:09.030
Finally, dividing both sides of our equation by negative two, we have 28 divided by negative two equals π₯.
00:02:09.350 --> 00:02:11.580
So π₯ equals negative 14.
00:02:11.970 --> 00:02:18.190
We can now plot these two coordinates on our equation two π₯ plus seven π¦ plus 28 equals zero.
00:02:18.540 --> 00:02:27.930
Weβre now asked to find the area of the triangle bounded by the π₯-axis, the π¦-axis, and the line two π₯ plus seven π¦ plus 28 equals zero.
00:02:28.180 --> 00:02:36.410
So we need to use the formula for the area of a triangle, which is that the area of a triangle is equal to half times the base times the height.
00:02:36.750 --> 00:02:44.200
So to find the area of our triangle, we could take the base to be the length along the π₯-axis, which is 14 units long.
00:02:44.540 --> 00:02:48.330
The height of the triangle can be taken as four units long.
00:02:48.780 --> 00:02:54.150
Therefore, we multiply a half, 14, and four, which gives us 28.
00:02:54.550 --> 00:03:06.030
Therefore, our final answer for the area of the triangle bounded by the π₯-axis, the π¦-axis, and the line with equation two π₯ plus seven π¦ plus 28 equals zero is 28 area units.