Collinearity of points in 2D geometry If the points with vertices $(p_1,q_1)$ , $(p_2,q_2)$ and $(p_1+p_2,q_1+q_2)$ are collinear , show that $p_1q_2=p_2q_1$
 A: The line through $(p_1,q_1)$ and $(p_2,q_2)$ is given by 
$$(p_1,q_1)+k(p_1-p_2,q_1-q_2),\quad \text{$k$ is a scalar number}$$
once $(p_1+p_2,q_1+q_2)$ is on the line then:
$$(p_1,q_1)+k(p_1-p_2,q_1-q_2)=(p_1+p_2,q_1+q_2)$$
what give us:
$$p_1+k(p_1-p_2)=p_1+p_2 \to k(p_1-p_2)=p_2\\
q_1+k(q_1-q_2)=q_1+q_2 \to k(q_1-q_2)=q_2$$
Isolating $k$ we have:
$$\frac{p_2}{p_1-p_2}=\frac{q_2}{q_1-q_2}\to p_2(q_1-q_2)=q_2(p_1-p_2)\\
p_2q_1-p_2q_2=p_1q_2-p_2q_2 \to p_2q_1=p_1q_2$$
A: Equating the slope of the lines between $(p_1,q_1), (p_2,q_2)$ and $(p_2,q_2), (p_1+p_2,q_1+q_2)\,$, respectively, gives (assuming $p_2 \ne p_1\ne 0$):
$$
\require{cancel}
\frac{q_2-q_1}{p_2-p_1} = \frac{(q_1+\bcancel{q_2})-\bcancel{q_2}}{(p_1+\cancel{p_2})-\cancel{p2}} \;\;\iff\;\; p_1(q_2-\cancel{q_1}) = (p_2-\cancel{p_1})q_1 \;\;\iff\;\; p_2q_1=p_1q_2
$$
The remaining (trivial) cases $p_2=p_1$ or $p_1=0$ need to be treated separately.
A: Let's try:
The equation of a straight line that passes through $(p_1,q_1)$ and $(p_2,q_2)$:
$y - q_1 = \frac { q_2 - q_1}{p_2 - p_1} ( x - p_1)$ .
(Standard formula).
This line also passes through$ (p_1 + p_2, q_1 + q_2)$.
Insert these values into the above equation of the line:
$q_1 + q_2 - q_1 = \frac{q_2 - q_1}{p_2 - p_1} ( p_1 +p_2 - p_1)$;
$q_2  = \frac{q_2 - q_1}{p_2 - p_1}(p_2)$;
$(p_2 - p_1) q_2 = p_2 ( q_2- q_1)$;
$p_1 q_2 = p_2 q_1$ .
