1.8 KiB
Analytical Geometry
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 = \(`\large (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} \large )`\)