Answer:
The coordinates for the location of the center of the platform are (-1 , 2)
Step-by-step explanation:
* Lets revise the equation of the circle
- The equation of the circle of center (h , k) and radius r is:
(x - h)² + (y - k)² = r²
- The center is equidistant from any point lies on the circumference
of the circle
- There are three points equidistant from the center of the circle
- We have three unknowns in the equation of the circle h , k , r
- We will substitute the coordinates of these point in the equation of
the circle to find h , k , r
* Lets solve the problem
∵ The equation of the circle is (x - h)² + (y - k)² = r²
∵ Points D (-2 , -4) , E (1 , 5) , F (2 , 0)
- Substitute the values of x and y b the coordinates of these points
# Point D (-2 , -4)
∵ (-2 - h)² + (-4 - k)² = r² ⇒ (1)
# Point E (1 , 5)
∵ (1 - h)² + (5 - k)² = r² ⇒ (2)
# Point (2 , 0)
∵ (2 - h)² + (0 - k)² = r²
∴ (2 - h)² + k² = r² ⇒ (3)
- To find h , k equate equation (1) , (2) and equation (2) , (3) because
all of them equal r²
∵ (-2 - h)² + (-4 - k)² = (1 - h)² + (5 - k)² ⇒ (4)
∵ (1 - h)² + (5 - k)² = (2 - h)² + k² ⇒ (5)
- Simplify (4) and (5) by solve the brackets power 2
# (a ± b)² = (a)² ± (2 × a × b) + (b)²
# Equation (4)
∴ [(-2)² - (2 × 2 × h) + (-h)²] + [(-4)² - (2 × 4 × k) + (-k)²] =
[(1)² - (2 × 1 × h) + (-h)²] + [(5)² - (2 × 5 × k) + (-k)²]
∴ 4 - 4h + h² + 16 - 8k + k² = 1 - 2h + h² + 25 - 10k + k² ⇒ add like terms
∴ 20 - 4h - 8k + h² + k² = 26 - 2h - 10k + h² + k² ⇒ subtract h² and k²
from both sides
∴ 20 - 4h - 8k = 26 - 2h - 10k ⇒ subtract 20 and add 2h , 10k
for both sides
∴ -2h + 2k = 6 ⇒ (6)
- Do the same with equation (5)
# Equation (5)
∴ [(1)² - (2 × 1 × h) + (-h)²] + [(5)² - (2 × 5 × k) + (-k)²] =
[(2)² - (2 × 2 × h) + k²
∴ 1 - 2h + h² + 25 - 10k + k² = 4 - 4h + k²⇒ add like terms
∴ 26 - 2h - 10k + h² + k² = 4 - 4h + k² ⇒ subtract h² and k²
from both sides
∴ 26 - 2h - 10k = 4 - 4h ⇒ subtract 26 and add 4h
for both sides
∴ 2h - 10k = -22 ⇒ (7)
- Add (6) and (7) to eliminate h and find k
∴ - 8k = -16 ⇒ divide both sides by -8
∴ k = 2
- Substitute this value of k in (6) or (7)
∴ 2h - 10(2) = -22
∴ 2h - 20 = -22 ⇒ add 20 to both sides
∴ 2h = -2 ⇒ divide both sides by 2
∴ h = -1
* The coordinates for the location of the center of the platform are (-1 , 2)
Answer:
The coordinates for the location of the center of the platform are (-3.5,1.5)
Step-by-step explanation:
You have 3 points:
D(−2,−4)
E(1,5)
F(2,0)
And you have to find a equidistant point (c) ([tex]x_{c}[/tex],[tex]y_{c}[/tex]) from the three given.
Then, you know that:
[tex]D_{cD}=D_{cE}[/tex]
And:
[tex]D_{cE}=D_{cF}[/tex]
Where:
[tex]D_{cD}[/tex]=Distance between point c to D
[tex]D_{cE}[/tex]=Distance between point c to E
[tex]D_{cF}[/tex]=Distance between point c to D
The equation to calculate distance between two points (A to B) is:
[tex]D_{AB}=\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2)}[/tex]
[tex]D_{AB}=\sqrt{(x_{B}^2)-(2*x_{B}*x_{A})+(x_{A}^2)+(y_{B}^2)-(2*y_{B}*x_{A})+(y_{A}^2)}[/tex]
Then you have to calculate:
*[tex]D_{cD}=D_{cE}[/tex]
[tex]D_{cD}=\sqrt{(x_{D}-x_{c})^2+(y_{D}-y_{c})^2}[/tex]
[tex]D_{cD}=\sqrt{(x_{D}^2)-(2*x_{D}*x_{c})+(x_{c}^2)+(y_{D}^2)-(2*y_{D} y_{c})+(y_{c}^2)}[/tex]
[tex]D_{cD}=\sqrt{(-2^2-(2(-2)*x_{c})+x_{c}^2)+(-4^2-(2(-4) y_{c})+y_{c}^2)}[/tex]
[tex]D_{cD}=\sqrt{(4+4x_{c}+x_{c}^2 )+(16+8y_{c}+y_{c}^2)}[/tex]
[tex]D_{cE}=\sqrt{(x_{E}-x_{c})^2+(y_{E}-y_{c})^2}[/tex]
[tex]D_{cE}=\sqrt{(x_{E}^2)-(2*x_{E}*x_{c})+(x_{c}^2)+(y_{E}^2)-(2y_{E}*y_{c})+(y_{c}^2)}[/tex]
[tex]D_{cE}=\sqrt{(1^2-2(1)*x_{c}+x_{c}^2)+(5^2-2(5)+y_{c}+y_{c}^2)}[/tex]
[tex]D_{cE}=\sqrt{(1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)}[/tex]
[tex]D_{cD}=D_{cE}[/tex]
[tex]\sqrt{((4+4x_{c}+x_{c}^2)+(16+8y_{c}+y_{c}^2))}=\sqrt{(1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)}[/tex]
[tex](4+4x_{c}+x_{c}^2)+(16+8y_{c}+y_{c}^2)= (1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2)[/tex]
[tex]x_{c}^2+y_{c}^2+4x_{c}+8y_{c}+20=x_{c}^2+y_{c}^2-2x_{c}-10y_{c}+26[/tex]
[tex]4x_{c}+2x_{c}+8y_{c}+10y_{c}=6[/tex]
[tex]6x_{c}+18y_{c}=6[/tex]
You get equation number 1.
*[tex]D_{cE}=D_{cF}[/tex]
[tex]D_{cE}=\sqrt{(x_{E}-x_{c})^2+(y_{E}-y_{c})^2}[/tex]
[tex]D_{cE}=\sqrt{(x_{E}^2-(2+x_{E}*x_{c})+x_{c}^2)+(y_{E}^2-(2y_{E} *y_{c})+y_{c}^2)}[/tex]
[tex]D_{cE}=\sqrt{((1^2-2(1)+x_{c}+x_{c}^2)+(5^2-2(5)y_{c}+y_{c}^2)}[/tex]
[tex]D_{cE}=\sqrt{(1-2x_{c}+x_{c}^2 )+(25-10y_{c}+y_{c}^2)}[/tex]
[tex]D_{cF}=\sqrt{(x_{F}-x_{c})^2+(y_{F}-y_{c})^2}[/tex]
[tex]D_{cF}=\sqrt{(x_{F}^2-(2*x_{F}*x_{c})+x_{c}^2)+(y_{F}^2-(2*y_{F}* y_{c})+y_{c}^2)}[/tex]
[tex]D_{cF}=\sqrt{(2^2-(2(2)x_{c})+x_{c}^2)+(0^2-(2(0)y_{c}+y_{c}^2)}[/tex]
[tex]D_{cF}=\sqrt{(4-4x_{c}+x_{c^2})+(0-0+y_{c}^2)}[/tex]
[tex]D_{cE}=D_{cF}[/tex]
[tex]\sqrt{(1-2x_{c}+x_{c}^2 )+(25-10y_{c}+y_{c}^2)}=\sqrt{(4-4x_{c}+x_{c}^2 )+(0-0+y_{c}^2)}[/tex]
[tex](1-2x_{c}+x_{c}^2)+(25-10y_{c}+y_{c}^2 )=(4-4x_{c}+x_{c}^2)+(0-0+y_{c}^2)[/tex]
[tex]x_{c}^2+y_{c}^2-2x_{c}-10y_{c}+26=x_{c}^2+y_{c}^2-4x_{c}+4[/tex]
[tex]-2x_{c}+4x_{c}-10y_{c}=-22[/tex]
[tex]2x_{c}-10y_{c}=-22[/tex]
You get equation number 2.
Now you have to solve this two equations:
[tex]6x_{c}+18y_{c}=6[/tex] (1)
[tex]2x_{c}-10y_{c}=-22[/tex] (2)
From (2)
[tex]-10y_{c}=-22-2x_{c}[/tex]
[tex]y_{c}=(-22-2x_{c})/(-10)[/tex]
[tex]y_{c}=2.2+0.2x_{c}[/tex]
Replacing [tex]y_{c}[/tex] in (1)
[tex]6x_{c}+18(2.2+0.2x_{c})=6[/tex]
[tex]6x_{c}+39.6+3.6x_{c}=6[/tex]
[tex]9.6x_{c}=6-39.6[/tex]
[tex]x_{c}=6-39.6[/tex]
[tex]x_{c}=-3.5[/tex]
Replacing [tex]x_{c}=-3.5[/tex] in
[tex]y_{c}=2.2+0.2x_{c}[/tex]
[tex]y_{c}=2.2+0.2(-3.5)[/tex]
[tex]y_{c}=2.2+0.2(-3.5)[/tex]
[tex]y_{c}=2.2-0.7[/tex]
[tex]y_{c}=1.5[/tex]
Then the coordinates for the location of the center of the platform are (-3.5,1.5)
