Location of vertex of triangle on Cartesian plane 
One triangle is drawn on the Cartesian plane. Its vertices are located on $[-2,-6]$,$[5,-6]$ and
  $[\frac{10}{3},d]$. Find the two possible positions of $d$ so that the triangle has an area of 35 square units.

So, obviously this triangle is scalene. Using Pick's theorem:
$i\frac{b}{2}-1=a$
where $i$ is the amount of interior points in the polygon in a dotted plane (the Cartesian plane if you think about it),$b$ is the amount of dotted points on the boundary and $a$ is the total area. So:
$i\frac{\geq7}{2}-1=35$
but where do I go from here. As far as I know, there is no way to find out the perimeter with only one measurement. Can anyone work this out?
P.S. Please solve this using techniques familiar to a Year 7-8, because I'm only an extension Yr 7 student. Also, I don't know the symbol for the'greater than or equal to" inequality sign in HTML or whatever, so anyone is free to fix it.  Thanks
 A: All you need to use is the formula for area of the triangle $P=\frac{ah}{2}$
Notice, that both vertices $A$ and $B$ lay on the same line $y=-6$. Vertice $C$ lay on the line $y=d$.
The height of this triangle is equal to the distance between lines $y=-6$ and $y=d$, ie. $h=|d-(-6)|=|d+6|$.
The length $a$ of the side $AB$ is $a=5-(-2)=7$.
The area of the triangle is $P=\frac{ah}{2}=37$, so you have to solve the equation:
$$\frac{7|d+6|}{2}=35$$
There will be two solutions:

$d=4$ and $d=-16$

EDIT
If you don't know the absolute value, then you can think about this in that way - there are two cases: $d<-6$, then $h=d+6$, or $d\geq -6$, then $h=-6-d$.
In first case you have to solve the equation $\frac{7(d+6)}{2}=35$ and check, if the obtained solution is smaller than $-6$.
Analogically in the second case
A: The given coordinates are [-2, -6], [5, -6] and $\left[ \frac{10}{3}, d \right]$. It is clear from the coordinates that the length of the base is $|-2-5| = 7$ units as the $y$-coordinate of the first two points are fixed at -6 and hence the side joining first two points will be parallel to the $x$-axis. Now assuming that the height of the triangle is $h$, the area can be computed as 
\begin{align}
& A = \dfrac{1}{2} \times b \times h \tag{where, $b$ is the base of the trianlge} \\
\implies &35 = \dfrac{1}{2} \times 7 \times h \\
\implies & h = \dfrac{35 \times 2}{7} = 10
\end{align}
Hence the height of the triangle is 10 units. Now since the height is calculated from a line $y = -6$, therefore the $y$-coordinate of the third point should be 10 units away from the line $y = -6$ which will be any point on the line $y = 4$ or the line $y = -16$. Furthermore, since the $x$-coordinate of the third point is given as $\dfrac{10}{3}$, hence the coordinate of the third vertex of the triangle is $\left[ \dfrac{10}{3}, 4\right]$ or $\left[ \dfrac{10}{3}, -16\right]$, i.e., $d = 4, -16$.
A: There is another really simple way to calculate this. 
 In a generic case, let's say that the coordinates of a triangle are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$. The area of the triangle is given by - 

$$\Delta = \frac12 * | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$$

Here, you may take $(x_1, y_1) = (-2, -6)$ ; $(x_2, y_2) = (5, -6)$ and $(x_3, y_3) = (\frac{10}{3}, d)$ and solve by plugging in their values in the above formula. 
 On solving, the two values of $d$ come out to be $-16$ and $4$ . 
