A slope of a line connecting two points #A(x_a,y_a)# and #B(x_b,y_b)# 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)#, three points #A, 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]}
