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