From 700d29c843165d967132a831f9d07348d0b44ec9 Mon Sep 17 00:00:00 2001 From: James Su Date: Mon, 13 Jan 2020 00:53:39 +0000 Subject: [PATCH] Update Unit 1: Analytical Geometry Part 1.md --- .../Unit 1: Analytical Geometry Part 1.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) 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. $`M_{AB} = (\dfrac{8+12}{2}, \dfrac{12+4}{2}) = (10, 8)`$