diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md
index 6aecda9..b210c3e 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Analytical Geometry.md	
@@ -4,10 +4,73 @@
 
 Lets first find each of the side lengths to determine if the triangle is **obtuse**, **acute** or scalene.
 
-$`AB = \sqrt{(-1-7)^2 + (5-2)^2} = \sqrt{64 + 9} = \sqrt{73}`$
+$`\overline{AB} = \sqrt{(-1-7)^2 + (5-2)^2} = \sqrt{64 + 9} = \sqrt{73}`$
 
-$`BC = \sqrt{(7-(-1))^2 + (2-(-4))^2} = \sqrt{64 + 36} = \sqrt{100} = 10`$
+$`\overline{BC} = \sqrt{(7-(-1))^2 + (2-(-4))^2} = \sqrt{64 + 36} = \sqrt{100} = 10`$
 
-$`AC = \sqrt{(-1-(-1))^2 + (5-(-4))^2} = \sqrt{0^2 + 9^2} = \sqrt{81} = 9`$
+$`\overline{AC} = \sqrt{(-1-(-1))^2 + (5-(-4))^2} = \sqrt{0^2 + 9^2} = \sqrt{81} = 9`$
 
-$`\since
\ No newline at end of file
+$`\because \overline{AB} =\not \overline{BC} =\not \overline{AC}`$
+
+$`\therefore \triangle ABC`$ is a scalene triangle.
+
+### Question 1 b)
+
+The `orthocenter` is the POI of the heights of a triangle.
+
+$`m_{AB} = \dfrac{2-5}{7-(-1)} = \dfrac{-3}{8}`$
+
+$`m_{\perp AB} = \dfrac{8}{3}`$
+
+$`y_{\perp AB} - (-4) = \dfrac{8}{3}(x - (-1)) \implies y_{perp AB} + 4 = \dfrac{8}{3}(x+1)`$
+
+$`y_{\perp AB} = \dfrac{8}{3}x + \dfrac{8}{3} - 4`$
+
+$` y_{\perp AB} = \dfrac{8}{3}x - \dfrac{4}{3} \quad (1)`$
+
+$`m_{BC} = \dfrac{2-(-4)}{7-(-1)} = \dfrac{6}{8} = \dfrac{3}{4}`$
+
+$`m_{\perp BC} = \dfrac{-4}{3}`$
+
+$`y_{\perp BC} - 5 = \dfrac{-4}{3}(x-(-1)) \implies y_{\perp BC} - 5 = \dfrac{-4}{3}(x+1)`$
+
+$`y_{\perp BC} = \dfrac{-4}{3}x - \dfrac{4}{3} + 5`$
+
+$`y_{\perp BC} = \dfrac{-4}{3}x + \dfrac{11}{3}`$
+
+```math
+
+\begin{cases}
+
+y_{\perp AB} = \dfrac{8}{3}x - \dfrac{4}{3} & \text{(1)} \\
+
+\\ 
+y_{\perp BC} = \dfrac{-4}{3}x + \dfrac{11}{3} & \text{(2)} \\
+\end{cases}
+```
+
+Sub $`(1)`$ into $`(2)`$:
+
+$`\dfrac{8}{3}x - \dfrac{4}{3} = \dfrac{-4}{3} + \dfrac{11}{3}`$
+
+$`8x - 4 = -4x + 11`$
+
+$`12x = 15`$
+
+$`x = \dfrac{5}{4} \quad (3)`$
+
+Sub $`(3)`$ into $`(2)`$
+
+$`y = \dfrac{-20}{12} + \dfrac{11}{3}`$
+
+$`y = \dfrac{-5}{3} + \dfrac{11}{3}`$
+
+$`y = \dfrac{6}{3} = 2`$
+
+$`y = 2`$
+
+$`\therefore`$ The `orthocenter` is at $`(\dfrac{5}{4}, 2)`$
+
+### Question 2 a)
+
+midpoint = $`\large (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} \large )`$