Finding a point in a parallelogram QUESTION:
Find the point$(x,y)$ so that $(x,y)$ is in the first quadrant and $(x,y),(1,2),(4,10)$ and $(2,6)$ are vertices of a parallelogram..
I find this question very difficult..
Thanks...
 A: A parallelogram has opposite sides that are equal and parallel. So you can say that the vectors representing opposite sides are the same (or one is $-1$ times the other).
$${x\choose y}-{1\choose2}={4\choose10}-{2\choose6}$$
Or if you wanted to use the other pair of sides
$${4\choose10}-{x\choose y}={2\choose6}-{1\choose2}$$
Either way, you will end up with the same answer.
A: As the diagonals of parallelogram bisect each other (Proof), 
the midpoint of $(x,y);(4,10)$ will be same as that of $(1,2);(2,6)$
A: I notice you have put this question in the trigonometry section, so I will answer using trigonometry rather than vectors, as someone else has, even though using vectors is probably the easiest way to solve the problem.
The first thing to do when solving these problems is to try and draw a graph.  I understand the graph is of an unusual parallelogram that looks almost like a rhombus, so drawing it may be difficult.
Remember that opposite sides of a parallelogram will have the same gradient.  So I'll work out the gradient of both sides, one of which will involve the unknown vertex coordinates.
Let $(x_1 , y_1) = (1 , 2)$
Let $(x_2 , y_2) = (2 , 6)$
Let $(x_3 , y_3) = (4 , 10)$
Let (x , y) = the coordinates of the unknown vertex.
To find the gradient use the gradient formula and substitute in the values.  
$$
\begin{align}
m &= \frac {y_2 - y_1} {x_2 - x_1} \\
  &= \frac {6 - 2} {2 - 1} \\
  &= \frac {4} {1} \\
  &= 4
\end{align}
$$
Now find the gradient of the opposite side, which will will involve the unknown vertex coordinates.  
\begin{align}
m_2 &= \frac {y_2 - y} {x_2 - x} \\
  &= \frac {10 - y} {4 - x} \\
\end{align}  
Because the gradients are on opposite sides and therefore the same, let
\begin{align}
m &= m_2 \\
m &= \frac {y_2 - y} {x_2 - x} \\
4 &= \frac {10 - y} {4 - x} \\
16 - 4x &= 10 - y \\
y &= 4x -6 \tag{1} \\
\end{align}  
Now do a similar process work out the gradients for the two other sides, which will involve the unknown (x,y) coordinates again.  Which will give the equation  $y=2x$  Then you'll have two equations that you can solve for x and y as simultaneous equations.  
