A slope of a line connecting two points A(x_a,y_a)A(xa,ya) and B(x_b,y_b)B(xb,yb) is a tangent of an angle from the positive direction of the X-axis counterclockwise to a line connecting these two points.
Together with a point C(x_b,y_a)C(xb,ya), three points A, B, CA,B,C form a right triangle DeltaABC with an angle /_BAC being exactly the one tangent of which we need.
The opposite to our angle side BC is measured as y_b-y_a and adjacent side AC equals to x_b-x_a.
Therefore, the tangent of angle BAC, that is the slope, equals to
tan(/_BAC)=(y_b-y_a)/(x_b-x_a)
In our case
x_a=3, y_a=4, x_b=6, y_b=9
Slope equals to
(9-4)/(6-3)=5/3
The graph of a line with a slope of 5/3 that passes through points A(3,4) and B(6,9) is below.
I recommend to mark on this graph all three points A(3,4), B(6,9) and C(6,4) and draw the right triangle DeltaABC.
graph{(5/3)x-1 [-1, 10, -2, 10]}
