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# triangle

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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
+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

Dec 29, 2017
#2
+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 $$\frac{1}{2}\cdot \frac{9}{5}\cdot 6=\boxed{\frac{27}{5}}$$

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Dec 29, 2017
#3
+4569
+1

Thanks, CPhill and azsun!

Dec 29, 2017