Find x-intercept of y=15 on huge ordered pair I've got the ordered pairs $(1491947996, 15.7)$ and $(1491948898, 12.9)$. The X values in each set indicate the seconds from Unix Epoch and the Y values in each set indicate the temperature in Celsius. 
I'm trying to determine the X-coordinate where $Y = 15$ (degrees Celsius), which will tell me the precise time in seconds when a temperature dropped below a safe-zone of 15°C. 
Naturally, $y = mx + b$, and $x = \frac{y-b}{m}$. I've got the slope, which is $0.003$, using $\frac{y2 - y1}{x2 - x1}$, or $\frac{12.9 - 15.7}{ 1491948898 - 1491947996}$
I guess what I don't know is how to define $y$ or $b$ in this case. I thought $y$ would certainly be $15$, but I guess I'm lost. It's been a long time since my last algebra course.
 A: What you want is called the "point-slope form" of a line. This is, for a given slope $m$ and a given point on the line $(x_0,~ y_0)$:
$$y-y_0 = m(x-x_0)$$
A: Algebraically, which means assuming your data points are exact, once you have $m$ you can just plug one of your points into $y=mx+b$ and find $b$.  Your slope has the wrong sign and is not exact.  Alpha gives about $-0.003104$.  Once you have $b$, you are correct and should substitute $y=15$ and solve for $x$.  
Alternately, once you have $m$ you can use $y-y_1=m(x-x_1)$.  Since your first point is close you can say $-0.7=-0.003104(x-1491947996)$ and find $x=1491948222$  
As you say, the huge $x$ numbers are scary.  If you believe they were accurately measured times, you don't have to worry about it.  It would be better to subtract $1491947996$ or some convenient number from both $x$ values.  That just represents changing the zero of time to a convenient time instead of using the Unix zero.  The scary part comes because you worry that you need such relative accuracy for the difference of the $x$ values to be significant, but what you really measured was the difference in the $x$ values, then you added a big number on to them.
A: You probably will find it easiest to use the two point form, since that's your starting point. Once you have the equation you can substitute 15 for y and get the value for X, in seconds since Jan 1, 1970. You really don't care about b, the temperature at the beginning of 1970.
A: 
Offset the measurements by
$$
 x_{0} = 1491947995
$$
The transformation is $\tilde{x} = x - x_{0}.$ (As noted by @Ross Millikan.)
Now the data looks like this:
$$ 
p = \left\{ 1, 15.7 \right\}, \qquad
q = \left\{ 903, 12.9 \right\} 
$$
The equation for a line in slope-intercept form:
$$
 y(x) = mx + b
$$
The slope is computed via
$$
  m =  \frac{p_{y} - q_{y}} {p_{x} - q_{x}}
$$
The intercept can be computed from either $p$ or $q$:
$$
m = p_{y} - m p_{x}
$$
To find the solution point, solve for $x$:
$$
 15 = m x + b
$$

Slope $m$

$$
m = \frac{p_{y} - q_{y}} {p_{x} - q_{x}} \approx -0.00310421286031042
$$
Intercept $b$
$$
m = p_{y} - m p_{x} \approx 15.70310421286031
$$
Solution
$$
 \color{red}{s} = \left\{ \frac{15 - b}{m}, 15 \right\} = \color{red}{\left\{ 226.5 , 15 \right\}}
$$
