diff --git a/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md b/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md
index fa76114..6810e98 100644
--- a/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md	
+++ b/Grade 10/Math/MPM2DZ/Unit 1: Analytical Geometry.md	
@@ -106,12 +106,9 @@ The orthocenter of a triangle is the common intersection of the 3 lines containi
 
 ## Ratios
 - To calculate each segment of the line given the ratio, the answer is simply 
-- $`(x_1 + \dfrac{x_2 - x_1}{r}, y_1 + \dfrac{y_2 - y1}{r})`$, where $`r, (x_1,y_1) (x_2,y_2)`$ are the  **total** ratio, first point and second point respectively.
-- Note that the above is for moving up a line. When moving down, we simply subtract like so:
-- $`(x_2 - \dfrac{x_2 - x_1}{r}, y_2 - \dfrac{y_2 - y1}{r})`$
-- For example, from a point like $`(2, 3)`$ to a point ($`5, 6)`$, and having a ratio of $`2:1`$ split at point $`P`$, the coordindates of point $`P`$ is simply 
-- $`(5 - \dfrac{5-2}{3}, 7 - \dfrac{6-3}{3})`$
-- Which is $`(4, 6)`$
+- $`(x_1 + \dfrac{p(x_2 - x_1)}{r}, y_1 + \dfrac{p(y_2 - y_1)}{r})`$, where $`r, (x_1,y_1) (x_2,y_2), p`$ are the  **total** ratio, first point, second point and the amount of steps respectively.
+- Note that the above is for moving up a line. When moving down from the upper point, we simply subtract like so:
+- $`(x_2 - \dfrac{p(x_2 - x_1)}{r}, y_2 - \dfrac{p(y_2 - y_1)}{r})`$
 
 ## Shortest Distance From Point To a Line
 - The shortest distance is always a straightline, thus, the shortest distance from a point to a line must be **perpendicular.**