Question 49238
Point P ( x=13, y=2) Point Q ( x1= -5, y1 = 26)

In y = mx + b equation Slope of Line PQ is :

m = Delta ( Y ) / Delta (x)
 
m = (y1 - y)/(x1-x) = (26-2) / (-5-13) = -(24/18)= -1.3333 ( Note : negative slope )

Therefore Slope m = -1.333

Now Constant b = Y intercept i.e., Point where line PQ crosses Y axis at x=0

y = mx +b
b = y - mx = y1-mx1 
b = 2 - (-1.333)(13) = 2 + 17.329 = 19.329 = 19.33( rounded off to 2 digits after decimal point)

Therefore : Foa a line PQ : Slope m = -1.333 and b = 19.33

Now Equation can be written in standard format as below :

y = mx + b 

y = -1.333x + 19.33

Thats all...
B.Shridhar