diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md
index 62de48b..f926330 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md	
@@ -22,7 +22,7 @@ $`m_{AB} = \dfrac{2-5}{7-(-1)} = \dfrac{-3}{8}`$
 
 $`m_{\perp AB} = \dfrac{8}{3}`$
 
-$`y_{\perp AB} - (-4) = \dfrac{8}{3}(x - (-1)) \implies y_{perp AB} + 4 = \dfrac{8}{3}(x+1)`$
+$`y_{\perp AB} - (-4) = \dfrac{8}{3}(x - (-1)) \implies y_{\perp AB} + 4 = \dfrac{8}{3}(x+1)`$
 
 $`y_{\perp AB} = \dfrac{8}{3}x + \dfrac{8}{3} - 4`$
 
@@ -87,7 +87,7 @@ $`\therefore`$ The midpoint is at $`(5 \sqrt{2}, -3\sqrt{3})`$
 
 Center of mass = centroid.
 
-Centroid = where all median lines of a trinagle intersect.
+Centroid = where all median lines of a triangle intersect.
 
 $`M_{AB} = (\dfrac{8+12}{2}, \dfrac{12+4}{2}) = (10, 8)`$