4.7 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 = \(` (\dfrac{\sqrt{72} + \sqrt{32}}{2}, \dfrac{-\sqrt{12} - \sqrt{48}}{2} )`\)
\(` = ( \dfrac{6\sqrt{2} + 4\sqrt{2}}{2}, \dfrac{-2\sqrt{3}, -4\sqrt{3}}{2}) `\)
\(` = 3\sqrt{2} + 2\sqrt{2}, -\sqrt{3} - 2\sqrt{3}`\)
\(` = (5\sqrt{2}, -3\sqrt{3})`\)
\(`\therefore`\) 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 \(`(\dfrac{22}{3}, 8)`\)
Question 3
Shortest distance = straight perpendicular line that connets \(`A`\) to a point on line \(`\overline{GH}`\)
\(`M_{GH} = \dfrac{42+30}{38 + 16} = \dfrac{72}{54} = \dfrac{4}{3}`\)
\(`M_{\perp GH} = \dfrac{-3}{4}`\)
\(`y_{\perp GH} - 32 = \dfrac{-3}{4}(x+16)`\)
\(`y_{\perp GH} = \dfrac{-3}{4}x + 20 \quad (1)`\)
\(`y_{GH} + 30 = \dfrac{4}{3}(x+16)`\)
\(`y_{GH} = \dfrac{4}{3}x - \dfrac{26}{3} \quad (2)`\)
\begin{cases}
y_{\perp GH} = \dfrac{-3}{4}x + 20 & \text{(1)} \\
\\
y_{GH} = \dfrac{4}{3}x - \dfrac{26}{3} & \text{(2}) \\
\end{cases}Sub \(`(1)`\) into \(`(2)`\)
\(`\dfrac{-3}{4}x + 20 = \dfrac{4}{3}x - \dfrac{26}{3}`\)
\(`-9x + (12)20 = 16x - 4(26)`\)
\(`25x = 344`\)
\(`x = \dfrac{344}{25} \quad (3)`\)
Sub \(`(3)`\) into \(`(1)`\)
\(`y = \dfrac{-3}{4}(\dfrac{344}{25}) + 20`\)
\(`y = \dfrac{-258}{25} + 20`\)
\(`y = \dfrac{-257}{25} + \dfrac{500}{25}`\)
\(`y = {242}{25}`\)
Distance \(`= \sqrt{(-16-\dfrac{344}{25})^2 + (32 - \dfrac{242}{25})^2} = 37.2`\)
\(`\therefore`\) The shortest length pipe is \(`37.2`\) units.
Question 4
Let \(`(x, y)`\) be the center of the circle, and \(`r`\) be the radius of the circle.
\begin{cases}
(x-4)^2 + (y-8)^2 = r^2 & \text{(1)} \\
(x-5)^2 + (y-1)^2 = r^2 & \text{(2)} \\
(x+2)^2 + y^2 = r^2 & \text{(3)} \\
\end{cases}Sub \(`(1)`\) into \(`(2)`\)
\(`x^2 - 8x + 16 + y^2 - 16y + 64 = x^2 - 10x + 25 + y^2 -2y + 1`\)
\(`-8x -16y + 80 = -10x - 2y + 26`\)
\(`2x - 14y = -54`\)
\(`x - 7y = -27 \quad (4)`\)
Sub \(`(2)`\) into \(`(3)`\)
\(`x^2 + 10x + 25 + y^2 - 2y + 1 = x^2 + 4x + 4 + y^2`\)
\(`10x - 2y +26 = 4x + 4`\)
\(`14x + 2y = 22`\)
\(`7x + y = 11`\)
\(`y = 11 - 7x \quad (5)`\)
Sub \(`(5)`\) into \(`(4)`\)
\(`x - 7(11-7x) = -27`\)
\(`x - 77+ 49x = 27`\)
\(`50x = 50`\)
\(`x = 1 \quad (6)`\)
Sub \(`(6)`\) into \(`(5)`\)
\(`y = 11 - 7(1)`\)
\(`y = 4 \quad (7)`\)
Sub \(`(6), (7)`\) into \(`(3)`\)
\(`(1+2)^2 + 4^2 = r^2`\)
\(`r^2 = 16 + 9`\)
\(`r^2 = 25`\)
\(`\therefore (x-1)^2 + (y-4)^2 = 25`\) is the equation of the circle.