+4622 Find the area of the triangle bounded by the \(y\)-axis and the lines \(y-3x=-2\) and \(3y+x=12\) .
+129899 Look at the graph here:
https://www.desmos.com/calculator/tleqe1iopp
We have a triangle
Let the base be formed by the segment connecting (0,-2) and (0, 4)
And the length of this base = 6 units
Now.....an altitude can be drawn from the point (1.8, 3.4) to the y axis
And this altitude is 1.8 units
So.....the area of this triangle is
(1/2)(base)(altitude) =
(1/2)(6)(1.8) =
.9 * 6 =
5.4 units^2
+199 To start, we can find the -intercept of each of these lines. Using this, we can calculate the length of that side of the triangle, and use it as a base. Letting x=0 in the first equation gives y=-2 as a y-intercept. Letting x=0 in the second equation gives \(3y=12\Rightarrow y=4\) as a y-intercept. Therefore, the triangle has a length of \(4-(-2)=6\) on the y-axis.
The height of the triangle will be equal to the -coordinate of the intersection of the two lines. So, we need to solve for x in the system: \(\begin{align*} y-3x&=-2\\ 3y+x&=12 \end{align*}\) Multiply the first equation by 3, then subtract the second equation as shown: -10x=-18Therefore, \(x=\frac{18}{10}=\frac{9}{5}\) . This is equal to the height of the triangle. The area will be \(\frac{1}{2}\cdot \frac{9}{5}\cdot 6=\boxed{\frac{27}{5}}\)
