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) $15$


Please show your work and explain.
 A: 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 $h = 8 \implies x = 3 + 8 = 11.$
A: Hinz Take the first twopoint as base line. It has length 6. Therefore the height must be 8.
A: 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 $x<3, 3(3-x)=24\implies -5$
