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129 lines
2.7 KiB
Markdown
129 lines
2.7 KiB
Markdown
## Analytical Geometry Part 1
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### Question 1 a)
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Lets first find each of the side lengths to determine if the triangle is **obtuse**, **acute** or scalene.
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$`\overline{AB} = \sqrt{(-1-7)^2 + (5-2)^2} = \sqrt{64 + 9} = \sqrt{73}`$
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$`\overline{BC} = \sqrt{(7-(-1))^2 + (2-(-4))^2} = \sqrt{64 + 36} = \sqrt{100} = 10`$
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$`\overline{AC} = \sqrt{(-1-(-1))^2 + (5-(-4))^2} = \sqrt{0^2 + 9^2} = \sqrt{81} = 9`$
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$`\because \overline{AB} =\not \overline{BC} =\not \overline{AC}`$
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$`\therefore \triangle ABC`$ is a scalene triangle.
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### Question 1 b)
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The `orthocenter` is the POI of the heights of a triangle.
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$`m_{AB} = \dfrac{2-5}{7-(-1)} = \dfrac{-3}{8}`$
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$`m_{\perp AB} = \dfrac{8}{3}`$
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$`y_{\perp AB} - (-4) = \dfrac{8}{3}(x - (-1)) \implies y_{\perp AB} + 4 = \dfrac{8}{3}(x+1)`$
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$`y_{\perp AB} = \dfrac{8}{3}x + \dfrac{8}{3} - 4`$
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$` y_{\perp AB} = \dfrac{8}{3}x - \dfrac{4}{3} \quad (1)`$
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$`m_{BC} = \dfrac{2-(-4)}{7-(-1)} = \dfrac{6}{8} = \dfrac{3}{4}`$
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$`m_{\perp BC} = \dfrac{-4}{3}`$
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$`y_{\perp BC} - 5 = \dfrac{-4}{3}(x-(-1)) \implies y_{\perp BC} - 5 = \dfrac{-4}{3}(x+1)`$
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$`y_{\perp BC} = \dfrac{-4}{3}x - \dfrac{4}{3} + 5`$
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$`y_{\perp BC} = \dfrac{-4}{3}x + \dfrac{11}{3}`$
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```math
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\begin{cases}
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y_{\perp AB} = \dfrac{8}{3}x - \dfrac{4}{3} & \text{(1)} \\
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\\
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y_{\perp BC} = \dfrac{-4}{3}x + \dfrac{11}{3} & \text{(2)} \\
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\end{cases}
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```
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Sub $`(1)`$ into $`(2)`$:
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$`\dfrac{8}{3}x - \dfrac{4}{3} = \dfrac{-4}{3} + \dfrac{11}{3}`$
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$`8x - 4 = -4x + 11`$
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$`12x = 15`$
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$`x = \dfrac{5}{4} \quad (3)`$
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Sub $`(3)`$ into $`(2)`$
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$`y = \dfrac{-20}{12} + \dfrac{11}{3}`$
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$`y = \dfrac{-5}{3} + \dfrac{11}{3}`$
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$`y = \dfrac{6}{3} = 2`$
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$`y = 2`$
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$`\therefore`$ The `orthocenter` is at $`(\dfrac{5}{4}, 2)`$
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### Question 2 a)
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midpoint = $` (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} )`$
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$` = ( \dfrac{6\sqrt{2} + 4\sqrt{2}}{2}, \dfrac{-2\sqrt{3}, -4\sqrt{3}}{2}) `$
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$` = 3\sqrt{2} + 2\sqrt{2}, -\sqrt{3} - 2\sqrt{3}`$
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$` = (5\sqrt{2}, -3\sqrt{3})`$
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$`\therefore`$ The midpoint is at $`(5 \sqrt{2}, -3\sqrt{3})`$
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### Question 2 b)
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Center of mass = centroid.
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Centroid = where all median lines of a triangle intersect.
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$`M_{AB} = (\dfrac{8+12}{2}, \dfrac{12+4}{2}) = (10, 8)`$
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$`m_{M_{AB} C} = \dfrac{8-8}{10-2} = 0`$
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$`y_{M_{AB} C} = 8 \quad (1)`$
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$`M_{BC} = (\dfrac{12+2}{2}, {8+4}{2}) = (7, 6)`$
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$`m_{M_{BC} A} = \dfrac{6-12}{7-8} = 6`$
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$`y_{M_{BC}A} - 12 = 6(x-8)`$
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$`y_{M_{BC}A} = 6x - 48 +12`$
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$`y_{M_{BC}A} = 6x - 36 \quad (2)`$
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```math
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\begin{cases}
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y_{M_{BC} A} = 8 & \text{(1)} \\
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y_{M_{BC} A} = 6x - 36 & \text{(2)} \\
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\end{cases}
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```
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Sub $`(1)`$ into $`(2)`$
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$`8 = 6x - 36`$
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$`6x = 44`$
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$`x = \dfrac{44}{6} = \dfrac{22}{3} \quad (3)`$
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By $`(1)`$, $`y=8`$.
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$`\therefore`$ The centroid is at $`(\dfrac{22}{3}, 8)`$
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