As we go from (3,7) to (6,9), x increases by 3 and y increases by 2. Thus, the gradient (slope) of the line connecting these two points is
m = rise / run = 2/3. Using the slope-intercept formula y = mx + b, we obtain
7 = (2/3)(3) + b, or 7 = 2 + b, so we see that b = 5 and y = (2/3)x + 5. The y-intercept is (0, 5).
Next we find the x-intercept. We set y = (2/3)x + 5 = to 0 and solve for x:
(2/3)x = -5, or (3/2)(2/3)x = -5(3/2), or x = -15/2, so that the x-intercept is
(-15/2, 0). This line intersects the x-axis at (-15/2, 0).
Now look at the segment of this line connecting (-15/2, 0) and (0, 5). Here x increases by 15/2 and y increases by 5, and so the tangent of the acute angle in question is
tan Ф = 5 / (15/2) = 10 / 15 = 2/3.
Using the inverse tangent function, we get Ф = arctan 2/3, or approx.
33.7 degrees.
I believe you meant "the acute angle it makes with the positive x-axis."
