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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! – Arturo Magidin May 17 '11 at 19:57
Is there a question here? – Arturo Magidin May 17 '11 at 19:59
Good for you. Do you have a question? Your answer is not exactly correct. – Ross Millikan May 17 '11 at 20:01
@Arturo: True but I presume she wanted to know how to proceed. – user9413 May 17 '11 at 20:01
@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 $8$. – André Nicolas May 17 '11 at 20:19

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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+1 $\ddot\smile$ – Babak S. Sep 16 '13 at 10:31

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. – user9413 May 17 '11 at 20:01
Last I checked, $\frac{8}{3}\neq\frac{133}{50}$. – Arturo Magidin May 17 '11 at 20:10
@Arturo: Where have i gone wrong. Don't know. Checked calculation and it seems correct. – user9413 May 17 '11 at 20:13
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. – Arturo Magidin May 17 '11 at 20:15
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. – Arturo Magidin May 17 '11 at 20:30

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