A University has three student residences which are located at points (−6, 12) (0,0) (3,9). The University wants to build a tennis court an equal distance from all three residences. Determine the coordinates of the tennis court. What is the distance from a residence to the court to the nearest 10th of a metre? (1unit=40m).

a.  (-3, 6)

b. 268.3 m

Step-by-step explanation:

a. Determine the coordinates of the tennis court.

Let (x,y) be the coordinates of the tennis court. Since we have equal distances from the tennis court to each residence, we have using the equation for distance between two points (x₁,y₁) and (x₂,y₂), d = √[(y₂ - y₁)² + (x₂ - x₁)²]. Using the first two residences, (-6, 12) and (0,0) as (x₂,y₂) and (x, y) = (x₁,y₁), So,

d = √[(12 - y)² + (-6 - x)²]

d = √[(12 - y)² + (6 + x)²]

d = √[(144 - 24y + y² + 36 + 12x + x²]

d = √x² + y² + 12x - 24y + 180

d² = x² + y² + 12x - 24y + 180  (1)

Also, with (x₂, y₂) = (0,0) we have

d = √[(0 - y)² + (0 - x)²]

d = √[( - y)² + (- x)²]

d = √[y² + x²]

d = √[x² + y²]

d² = x² + y²  (2)

Also, with (x₂, y₂) = (3,9) we have

d = √[(9 - y)² + (3 - x)²]  

d = √[(81 - 18y + y² + 9 - 6x + x²]

d = √(x² + y² - 6x - 18y + 90)

d² = x² + y² - 6x - 18y + 90 (3)

Substituting (2) into (1), we have

d² = x² + y² + 12x - 24y + 180   (1)

d² = d² + 12x - 24y + 180  (1)

d² - d² = 12x - 24y + 180

12x - 24y + 180 = 0

12x - 24y = -180

dividing through by 12, we have

x - 2y = -15  (4)

Substituting (2) into (3), we have

d² = x² + y² - 6x - 18y + 90 (3)

d² = d² - 6x - 18y + 90 (3)

d² - d² = - 6x - 18y + 90

- 6x - 18y + 90 = 0

- 6x - 18y = -90

dividing through by -6, we have

x + 3y = 15 (5)

Subtracting (5) from (4), we have

x - 2y = -15  (4)

-

x + 3y = 15 (5)

-5y = -30

y = -30/-5

y = 6

substituting y = 6 into (4), we have

x - 2y = -15

x - 2(6) = -15

x - 12 = -15

x = 12 - 15

x = -3

So, the coordinates of the tennis court is (-3, 6)

b. What is the distance from a residence to the court to the nearest 10th of a metre? (1unit=40m).

Using equation (2),

d² = x² + y²  (2) where x and y are the coordinates of the tennis court, and x = -3 and y = 6

So,

d² = x² + y²  

d² = (-3)² + (6)²  

d² = 9 + 36  

d² = 45

d = √45 units

Since 1 unit = 40 m,

d = √45 × 40 m

d = 6.708 × 40 m

d = 268.33 m

d ≅ 268.3 m to the nearest tenth of a metre

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!