# Finding the intersection of a two points and an arbitrary axis

Given two points I would like to find where the line joining them intersects an arbitrary axis. For example, if I had one point (5, 10) and another at (50, 100) I can be sure that somewhere a line joining these two points will intersect a line running parallel to the x axis with the y value of 23. Is it possible to determine the x value at this point of intersection?

-

$$y=mx+b.\qquad (1)$$

If the line passes through $P_{1}(x_{1},y_{1})$, then

$$y_{1}=mx_{1}+b\qquad (2)$$

and

$$b=y_{1}-mx_{1}.\qquad (3)$$

If it passes through $P_{2}(x_{2},y_{2})$, then

$$y_{2}=mx_{2}+b.\qquad (4)$$

The difference $(4)-(2)$ gives

$$y_{2}-y_{1}=m(x_{2}-x_{1}).\qquad (5)$$

Hence $(1)$ becomes

$$y=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}x+y_{1}-\frac{y_{2}-y_{1}}{x_{2}-x_{1}}% x_{1}.\qquad (6)$$

For $P_{1}(5,10),P_{2}(50,100)$, the equation $(5)$ is

$$y=\frac{100-10}{50-5}x+10-\frac{100-10}{50-5}5,\qquad (7)$$

which is equivalent to

$$y=2x.\qquad (8)$$

For $y=23$, you have

$$23=2x.\qquad (9)$$

Thus

$$x=\frac{23}{2}.\qquad (10)$$

-

You can use the two point form of the equation for a line. For your example, it is $\frac{y-10}{x-5}=\frac{100-10}{50-5}$ Your line parallel to the $x$ axis is $y=23$. So plug that into the first equation and solve for $x$.

-