I think you're over-thinking it. When you have the intersect point and another point, you just duplicate the difference to get the point on the other side.
$(\frac{6}{5}, -\frac{3}{5}) - (4, -2)$ is $(-\frac{14}{5}, \frac{7}{5})$
That is what you need to add to the point to get to your intersect point, so either add that do your intersect point, or double it and add it to the initial point to get the point on the other side:
$(-\frac{8}{5}, \frac{4}{5})$
Another option would just be to calculate the x difference and double it to get the opposite x coordinate and pop that into your perpendicular line equation.
Full:
Initial equation: $y = 2x - 3$
Point: $(4, -2)$
So any equation perpendicular to the initial one will have slope $-\frac{1}{2}$ as you suggest. You might have already found the full equation for the line to find your point, but plug in the $y$ value for the given point of -2 and solve for b:
$y = -\frac{1}{2}x$ + b
$b=y +\frac{1}{2}x$
$b=-2 + \frac{1}{2}4$
$b= 0$
So the formula for the perpendicular line passing through point $(4, -2)$ is $y = -\frac{1}{2}x$
Solving to find the intersection point to get the $x$ you set both equations to be equal and solve for x:
$2x - 3 = -\frac{1}{2}x$
$\frac{5x}{2} = 3$
$5x = 6$
$x = \frac{6}{5}$
Plugging that into either equation gives your intersect point:
$(\frac{6}{5},-\frac{3}{5})$
Ok, so now how do you find a point the same distance on the perpendicular line passing through $(4, -2)$? To get from $(4, -2)$ to $(\frac{6}{5}, -\frac{3}{5})$ you have to move $\frac{6}{5} - 4$ in the x direction and $-\frac{3}{5} - -2$ in the y direction, or $(-\frac{14}{5}, \frac{7}{5})$
Add that to the intersect point and you get the point on the opposite side, $(-\frac{8}{5}, \frac{4}{5})$. Plug the x value into your perpendicular equation:
$y = -\frac{1}{2}x$
$y = (-\frac{1}{2}) (-\frac{8}{5}$)
$y = \frac{8}{10}$
$y = \frac{4}{5}$
What you have defined is called a metric on $\mathbb{R}^4$ - a notion of a distance function for points in four-dimensional real space. Mathematicians often work with the concept of metric spaces, formally defined as a pair $(E,d)$, where $E$ is a set and $d:E\times E\to\mathbb{R}$ satisfying a particular set of identities (where $x,y\in E$):
- $d(x,y)\geq0$
- $d(x,y)=0\iff x=y$
- $d(x,y)=d(y,x)$
- $d(x,z)\leq d(x,y)+d(y,z)$
So, in your case, what you have conjectured is that the function $d:\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}$ satisfying:
$$d\big((x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4)\big)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2+(x_4-y_4)^2}$$
is a metric on $\mathbb{R}^4$. Try proving it!
But there are other metrics on $\mathbb{R}^4$ other than just the Euclidean one. For example, there is the taxicab metric:
$$d\big((x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4)\big)=\sum_{i=1}^4|x_i-y_i|$$
And also the discrete metric:
$$d(\vec x,\vec y)=
\begin{cases}
1, & \vec x\neq\vec y\\
0, & \vec x=\vec y \\
\end{cases}$$
It is also a good exercise to prove that these are also metrics on $\mathbb{R}^4$, and hence also perfectly valid notions of distance according to our definition. Your extension of the Euclidean metric to $\mathbb{R}^4$ preserves some nice qualities that we are used to - namely, translation and rotation invariance - but it is good to know that it is not the only way to define distance in $\mathbb{R}^4$.
As for how to interpret points in $\mathbb{R}^4$, think about how you view points in lower dimensions first. In $\mathbb{R}$, points lie on a number line. In $\mathbb{R}^2$, points lie on a plane. In $\mathbb{R}^3$, points lie in a larger three-dimensional space. In each case, we add a new dimension to where we can consider a point lying, but fundamentally nothing much changes about how we think about and consider points. You could, if you wanted to, just think about a point in $n$-dimensional real space as an ordered $n$-tuple of coordinates, each lying somewhere on the real line $\mathbb{R}$. So taking the case $n=4$, we can think of a point in $\mathbb{R}^4$ as being an ordered $4$-tuple of coordinates $(x_1,x_2,x_3,x_4)$ each lying on the real number line somewhere.
Best Answer
It could be useful if you had the length of the hypotenuse (d or the length of the line segment) and the distance that the segment covers on the x-axis $(x_2 - x_1)$ and you wanted to find the positive slope.
Ex: You have a segment that is 5 units long and covers 4 units on the x-axis. Using your equation, the positive slope of the line would be $\frac{3}{4}$.