2.0 KiB
Analytical Geometry Part 2
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 \(`(1)`\)
\(`y = \dfrac{-3}{4}(\dfrac{344}{25}) + 20`\)
\(`y = \dfrac{-258}{25} + 20`\)
\(`y = \dfrac{-258}{25} + \dfrac{500}{25}`\)
\(`y = \dfrac{242}{25}`\)
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.