Variance of two weighted random variables Let:
Standard deviation of random variable $A =\sigma_{1}=5$
Standard deviation of random variable $B=\sigma_{2}=4$
Then the variance of A+B is:
$Var(w_{1}A+w_{2}B)= w_{1}^{2}\sigma_{1}^{2}+w_{2}^{2}\sigma_{2}^{2} +2w_{1}w_{2}p_{1,2}\sigma_{1}\sigma_{2}$
Where:
$p_{1,2}$ is the correlation between the two random variables.
$w_{1}$ is the weight of random variable A
$w_{2}$ is the weight of random variable B
$w_{1}+w_{2}=1$
The figure below plots the variance of A and B as the weight of A changes from 0 to 1, for the correlations -1 (yellow),0 (blue) and 1 (red).

How did the formula result in a straight line (red) when the correlation is 1? As far as I can tell, when $p_{1,2}=1$, the formula simplifies to:
$Var(w_{1}A+w_{2}B)= w_{1}^{2}\sigma_{1}^{2}+w_{2}^{2}\sigma_{2}^{2} +2w_{1}w_{2}\sigma_{1}\sigma_{2}$
How can I express that in the form of $y=mx+c$?
Thank you.
 A: Using $w_1 + w_2 = 1$, compute
$$\eqalign{
\text{Var}(w_1 A + w_2 B) &= \left( w_1 \sigma_1 + w_2 \sigma_2 \right)^2 \cr
&= \left( w_1(\sigma_1 - \sigma_2) + \sigma_2 \right)^2 \text{.}
} $$
This shows that when $\sigma_1 \ne \sigma_2$, the graph of the variance versus $w_1$ (shown sideways in the illustration) is a parabola centered at $\sigma_2 / (\sigma_2 - \sigma_1)$.  No portion of any parabola is linear.  With $\sigma_1 = 5$ and $\sigma_2 = 4$, the center is at $-5$: way below the graph at the scale in which it is drawn.  Thus, you are looking at a small piece of a parabola, which will appear linear.
When $\sigma_1 = \sigma_2$, the variance is a linear function of $w_1$.  In this case the plot would be a perfectly vertical line segment.
BTW, you knew this answer already, without calculation, because basic principles imply the plot of variance cannot be a line unless it is vertical.  After all, there is no mathematical or statistical prohibition to restrict $w_1$ to lie between $0$ and $1$: any value of $w_1$ determines a new random variable (a linear combination of the random variables A and B) and therefore must have a non-negative value for its variance.  Therefore all these curves (even when extended to the full vertical range of $w_1$  ) must lie to the right of the vertical axis.  That precludes all lines except vertical ones.
Plot of the variance for $\rho = 1 - 2^{-k}, k = -1, 0, 1, \ldots, 10$:

A: It isn't linear.  The formula says it isn't linear.  Trust your mathematical instinct!  
It only appears linear in the graph because of the scale, with $\sigma_{1}=5$ and $\sigma_{2}=4$. Try it yourself: calculate the slopes at a few places and you will see that they differ.  You can exaggerate the difference by picking $\sigma_{1}=37$, say.
Here is some R code:
a <- 5; b <- 4; p <- 1
f <- function(w) w^2*a^2 + (1-w)^2*b^2 + 2*w*(1-w)*p*a*b
curve(f, from = 0, to = 1)

If you would like to check some slopes:
(f(0.5) - f(0.4)) / 0.1
(f(0.8) - f(0.7)) / 0.1

