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Find the area of the triangle bounded by the \(y\)-axis and the lines \(y-3x=-2\) and \(3y+x=12\) .

 Dec 29, 2017
 #1
avatar+111329 
+1

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

 

 

cool cool cool

 Dec 29, 2017
 #2
avatar+191 
+2

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 \(\frac{1}{2}\cdot \frac{9}{5}\cdot 6=\boxed{\frac{27}{5}}\)

.
 Dec 29, 2017
 #3
avatar+4569 
+1

Thanks, CPhill and azsun!

 Dec 29, 2017

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