Suppose I have 3 points. A and B and a C. I want the line segment throu C, parallel to the line on which A and B are.
In the image above I want the A' and B' points.
I know I can get the slope of the line between A and B and get the line passing through C with the y = mx + b formula. But I'm not sure how to get the A' and B' points.
Edit
Sorry I didn't give any information about the A' and B' points. Basically they're the same points on the new line segment if we map every point to the new line. This means that A' is a point on the new line with the same distance as C from AB's line. Sorry if there's a word in geometry for this. English is not my native language.
3 Answers
$\begingroup$The equation of a line through $(x_A,y_A)$ and $(x_B,y_B)$ is
$$A \mathbf{x} + B \mathbf{y} + C=0\\(y_B-y_A) \mathbf{x} + (x_A-x_B) \mathbf{y} + x_B y_A - x_A y_B = 0 $$
The equation of parallel passing through $(x_C,y_C)$ is
$$A \mathbf{x} + B \mathbf{y} + C'=0 \\(y_B-y_A) \mathbf{x} + (x_A-x_B) \mathbf{y} + y_C (x_B-x_A) + x_C (y_A-y_B) = 0 $$
The point on the parallel a line $A \mathbf{x} + B \mathbf{y} + C' = 0$ closest to point $(x_A,y_A)$ is
$$ (x_A',y_A') = \left( \frac{B^2 x_A-A (B y_A+C')}{A^2+B^2},\; \frac{A^2 y_A-B (A x_A+C')}{A^2+B^2} \right) $$
and similarly for the other point
$$ (x_B',y_B') = \left( \frac{B^2 x_B-A (B y_B+C')}{A^2+B^2},\; \frac{A^2 y_B-B (A x_B+C')}{A^2+B^2} \right) $$
with
$$\begin{align} A & = y_B-y_A = -13\\ B & = x_A-x_B = -25 \\ C & = x_B y_A - x_A y_B = 630 \\ C' & = y_C (x_B-x_A) + x_C (y_A-y_B) = 406 \end{align} $$
$$ (x_A',y_A') = \left( \frac{2514}{397}, \; \frac{5140}{397} \right) = (6.33, 12.95) $$ $$ (x_B',y_B') = \left( \frac{12439}{397}, \; -\frac{21}{397} \right) = (31.33, -0.05)$$
Confirmation with Geogebra
After getting green line parallel to blue line containing $A,B$ you have freedom for $A'$ or $ B'$ to move anywhere on green line.
You have not given any clue how to fix $A'$ or $ B'$ locations so their exact position cannot be determined.
EDIT1
Add 2 and subtract 10 for $ x,y $ co-ordinates respectively to come from A to A' or B to B' in new positions.
So A goes from $(10,20) $ to $(12,10) $ and B goes from $(35,7) $ to $( 37, -3). $
It is called parallel displacement or translation of straight line AB to A'B' in new position. The lines $ AB, A'B', AA', BB' $ form a $ parallellogram.$
If you understood it please include the new position of B' in the new sketch. Present indication of B' is incorrect.
$\endgroup$ 3 $\begingroup$Notice,
the equation of line segment $A'B'$ passing through the point $C(12, 10)$ & parallel to the line passing through the points $A(10, 20)$ & $B(35, 7)$ is given as $$y-10=\frac{20-7}{10-35}(x-12)$$ $$y=\frac{-13}{25}x+\frac{406}{25}$$ Now, in order to get the foots of perpendicular points $A'$ & $B'$ from the points $A(10, 20)$ & $B(35, 7)$ on the line $A'B'$ Let the point $A'$ be $(a, b)$. Since the line $A'B'$ $\perp$ $AA'$ hence $$\frac{b-20}{a-10}\times \frac{-13}{25}=-1$$ $$25a-13b=-10\tag 1$$ Since, the point $A'(a, b)$ lies on the line $A'B'$ hence satisfying the equation of line $A'B'$ by the point $A'(a, b)$ we get $$13a+25b=406\tag 2$$ Now, solving (1) & (2), we get $$a=\frac{2514}{397}, \ b=\frac{5140}{397}$$ Hence, the point $A'$ is $\color{red}{\left(\frac{2514}{397}, \frac{5140}{397}\right)}$
I hope you can find out the coordinates of point $B'$ which is the reflection of point $B(35, 7)$ by following the similar method
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