# How do you find the slope of (3,4), (6,9)?

May 6, 2015

A slope of a line connecting two points $A \left({x}_{a} , {y}_{a}\right)$ and $B \left({x}_{b} , {y}_{b}\right)$ 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 $C \left({x}_{b} , {y}_{a}\right)$, three points $A , B , C$ form a right triangle $\Delta A B C$ with an angle $\angle B A C$ being exactly the one tangent of which we need.

The opposite to our angle side $B C$ is measured as ${y}_{b} - {y}_{a}$ and adjacent side $A C$ equals to ${x}_{b} - {x}_{a}$.
Therefore, the tangent of angle $B A C$, that is the slope, equals to
$\tan \left(\angle B A C\right) = \frac{{y}_{b} - {y}_{a}}{{x}_{b} - {x}_{a}}$

In our case
${x}_{a} = 3 , {y}_{a} = 4 , {x}_{b} = 6 , {y}_{b} = 9$

Slope equals to
$\frac{9 - 4}{6 - 3} = \frac{5}{3}$

The graph of a line with a slope of $\frac{5}{3}$ that passes through points $A \left(3 , 4\right)$ and $B \left(6 , 9\right)$ is below.
I recommend to mark on this graph all three points $A \left(3 , 4\right)$, $B \left(6 , 9\right)$ and $C \left(6 , 4\right)$ and draw the right triangle $\Delta A B C$.
graph{(5/3)x-1 [-1, 10, -2, 10]}