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

tertre  Dec 29, 2017

Look  at the graph here:




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

CPhill  Dec 29, 2017

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

azsun  Dec 29, 2017

Thanks, CPhill and azsun!

tertre  Dec 29, 2017

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