# The area of the triangle with vertices (3, 2), (3, 8), and (x, y) is 24. What is x?

The area of the triangle with vertices (3, 2), (3, 8), and (x, y) is 24. A possible value for x is: a) 7 b) 9 c) 11 d) 13 e) 15

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is it a homework? –  Yimin Feb 15 '13 at 3:44
It's for an upcoming test. But I don't understand. –  user62366 Feb 15 '13 at 3:48
Graph it. It should be obvious. –  Tpofofn Feb 15 '13 at 4:12

One side of the triangle lies on the line $x = 3$, and is length $6$. Why?. Take that to be your base, $b$.

The area of a triangle is given by $$\text{Area}\;=\;\dfrac 12 bh$$ where $h$ is the height of the triangle measured from the base (connecting the third point perpendicular to the base, so $$\dfrac12(6)h = 24 \iff h = 8$$

Now, height, h, is the perpendicular distance from the base, which is on $x = 3$, and the only possible choices for $x$ that are given are all positive.

Hence $h = 8 \implies x = 3 + 8 = 11.$

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Thank you! This helped. But would this way only work with right angle triangles? –  user62366 Feb 15 '13 at 4:07
Ok.But what if the one side of the triangle did not lie on the same line? –  user62366 Feb 15 '13 at 4:21
It doesn't work only if the triangle is a right triangle: We don't know yet what y might be, or where the perpendicular line from $(x,y)$ intersects the line x=3, where the base lies (the perpendicular distance from the point (x, y) to the which is height), only that it must intersect the line $x = 3$. It would be a right triangle iff y = 2 or y = 8, $y$ being the value of the unknown point. –  amWhy Feb 15 '13 at 14:10
We only know that the unknown point $x, y$ is limited by a perpendicular distance of $8$ to the line $x = 3$, and that the point is somewhere to the right of the line $x = 3$, because the choices for the unknown x value are all positive and $> 3$. –  amWhy Feb 15 '13 at 14:13

Hinz Take the first twopoint as base line. It has length 6. Therefore the height must be 8.

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We know the area of a triangle (Article#25) having vertices $(x_i,y_i)$ for $i=1,2,3$ is

$$\frac12\det\begin{pmatrix} x_1 & y_1 & 1\\x_2&y_2&1\\ x_3 & y_3 &1\end{pmatrix}$$

Now, $$\det\begin{pmatrix}x & y &1\\ 3 & 2 & 1 \\ 3 & 8 & 1 \end{pmatrix}$$

$$=\det\begin{pmatrix}x-3 & y-2 &0\\ 3 & 2 & 1 \\ 0 & 6 & 0 \end{pmatrix}\text { (Applying } R_3'=R_3-R_2,R_1'=R_1-R_2)$$

$$=6(x-3)$$

As we take the area in absolute value,the are here will be $\frac12\cdot6|x-3|=3|x-3|$

If $x\ge 3, 3(x-3)=24\implies x=11$

If $x<3, 3(3-x)=24\implies -5$

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