highschool/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 1: Analytical Geometry Part 1.md

Analytical Geometry Part 1

Question 1 a)

Lets first find each of the side lengths to determine if the triangle is obtuse, acute or scalene.

\(`\overline{AB} = \sqrt{(-1-7)^2 + (5-2)^2} = \sqrt{64 + 9} = \sqrt{73}`\)

\(`\overline{BC} = \sqrt{(7-(-1))^2 + (2-(-4))^2} = \sqrt{64 + 36} = \sqrt{100} = 10`\)

\(`\overline{AC} = \sqrt{(-1-(-1))^2 + (5-(-4))^2} = \sqrt{0^2 + 9^2} = \sqrt{81} = 9`\)

\(`\because \overline{AB} =\not \overline{BC} =\not \overline{AC}`\)

\(`\therefore \triangle ABC`\) is a scalene triangle.

Question 1 b)

The orthocenter is the POI of the heights of a triangle.

\(`m_{AB} = \dfrac{2-5}{7-(-1)} = \dfrac{-3}{8}`\)

\(`m_{\perp AB} = \dfrac{8}{3}`\)

\(`y_{\perp AB} - (-4) = \dfrac{8}{3}(x - (-1)) \implies y_{perp AB} + 4 = \dfrac{8}{3}(x+1)`\)

\(`y_{\perp AB} = \dfrac{8}{3}x + \dfrac{8}{3} - 4`\)

\(` y_{\perp AB} = \dfrac{8}{3}x - \dfrac{4}{3} \quad (1)`\)

\(`m_{BC} = \dfrac{2-(-4)}{7-(-1)} = \dfrac{6}{8} = \dfrac{3}{4}`\)

\(`m_{\perp BC} = \dfrac{-4}{3}`\)

\(`y_{\perp BC} - 5 = \dfrac{-4}{3}(x-(-1)) \implies y_{\perp BC} - 5 = \dfrac{-4}{3}(x+1)`\)

\(`y_{\perp BC} = \dfrac{-4}{3}x - \dfrac{4}{3} + 5`\)

\(`y_{\perp BC} = \dfrac{-4}{3}x + \dfrac{11}{3}`\)

\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 = \(` (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} )`\)

\(` = ( \dfrac{6\sqrt{2} + 4\sqrt{2}}{2}, \dfrac{-2\sqrt{3}, -4\sqrt{3}}{2}) `\)

\(` = 3\sqrt{2} + 2\sqrt{2}, -\sqrt{3} - 2\sqrt{3}`\)

\(` = (5\sqrt{2}, -3\sqrt{3})`\)

\(`\therefore`\) The midpoint is at \(`(5 \sqrt{2}, -3\sqrt{3})`\)

Question 2 b)

Center of mass = centroid.

Centroid = where all median lines of a trinagle intersect.

\(`M_{AB} = (\dfrac{8+12}{2}, \dfrac{12+4}{2}) = (10, 8)`\)

\(`m_{M_{AB} C} = \dfrac{8-8}{10-2} = 0`\)

\(`y_{M_{AB} C} = 8 \quad (1)`\)

\(`M_{BC} = (\dfrac{12+2}{2}, {8+4}{2}) = (7, 6)`\)

\(`m_{M_{BC} A} = \dfrac{6-12}{7-8} = 6`\)

\(`y_{M_{BC}A} - 12 = 6(x-8)`\)

\(`y_{M_{BC}A} = 6x - 48 +12`\)

\(`y_{M_{BC}A} = 6x - 36 \quad (2)`\)

\begin{cases}

y_{M_{BC} A} = 8 & \text{(1)} \\

y_{M_{BC} A} = 6x - 36 & \text{(2)} \\

\end{cases}

Sub \(`(1)`\) into \(`(2)`\)

\(`8 = 6x - 36`\)

\(`6x = 44`\)

\(`x = \dfrac{44}{6} = \dfrac{22}{3} \quad (3)`\)

By \(`(1)`\), \(`y=8`\).

\(`\therefore`\) The centroid is at \(`(\dfrac{22}{3}, 8)`\)