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A triangle has a vertex $A$ at $(0,3)$, vertex $B$ at $(4,0)$, and vertex $C$ at $(x,5)$. If the area of the triangle is $8$, what is the value of $x$?

I did this problem out by using the formula of the given coordinates of the vertices. My answer was 2.66.

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    Please put the problem **in the body of the question**, not in the title!2011-05-17
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    Is there a question here?2011-05-17
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    Good for you. Do you have a question? Your answer is not exactly correct.2011-05-17
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    @Arturo: True but I presume she wanted to know how to proceed.2011-05-17
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    or may be even check the answer with us.2011-05-17
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    @Chandru: I'm not criticising you, so I don't see why the need to defend yourself. But since you bring it up, why did you assume "she wanted to know how to proceed", if she said she had already finished?2011-05-17
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    @Arturo: Don't know. Can't really justify that.2011-05-17
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    @Briana791: If the question asked for the *positive* $x$ that works, you have found it. More precisely, $x=8/3$. But, as a picture may persuade you, there is also a negative $x$ that gives area $8$.2011-05-17

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Good work, in arriving at one of the correct solutions, Briana..

I would just like to clarify what others mean by there being two solutions: there are two triangles, each sharing vertices at $(0,3)$ and $(4,0)$ (and hence they share a side). But there are two values for the unknown $x$ in the vertex $(x, 5)$ that yield a triangle with area $8: x = \frac {8}{3} = (2.66$ repeating), and $x = -8.$

In other words, the area of the triangle with third vertex $( \frac {8}{3}, 5)$ and the area of triangle with third vertex $(-8, 5)$ both equal $8$.
two triangles, area 8

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The area of the triangle with vertices $(x_{i},y_{i})$, $i=1,2,3$ is given by $$\Delta = \frac{1}{2} \cdot \left| \begin{array}{cc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1 \end{array}\right| = \frac{1}{2}\left|\begin{array}{cc} 0 & 3 & 1 \\ 4 & 0 & 1 \\ x & 5 & 1 \end{array}\right|$$

Now substitute $\Delta=8$, and the corresponding values for $(x_{i},y_{i})$ and solve for $x$. Solving you get $16 = -3(4-x) + 1 \times 20 = -12 + 3x +20$. Therefore you have $3x=8$ which says $x=\frac{\pm{8}}{3}$. You get the minus sign if you interchange the vertices. (Thanks to Arturo for pointing out the error.)

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    @Briana: Does this clarify your doubts.2011-05-17
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    Last I checked, $\frac{8}{3}\neq\frac{133}{50}$.2011-05-17
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    @Arturo: Where have i gone wrong. Don't know. Checked calculation and it seems correct.2011-05-17
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    Sigh. $\frac{8}{3}\neq \frac{133}{50} = \frac{266}{100} = 2.66$. If you are giving approximations, don't use the equal sign. If you use the equal sign, both sides should be **equal**.2011-05-17
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    @Arturo: Thanks. I have edited.2011-05-17
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    The formula cannot be correct as given, because if you exchange the first two rows (if you take the vertices in a different order), then the value of the determinant is multiplied by $-1$, yielding a "negative area". So you are missing an absolute value sign, which also means that there is more than one solution.2011-05-17
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    @Chandru: As Arturo points out as well, the area is the magnitude of the determinant and hence you will get $x = -8$ as well.2011-05-17