A 30-lb child is siting at A(2,3) and a 50-lb child is at B(12,7) where units are feet .Find the point P between A and B which could be used as the fulcrum of a teeter board putting the two children in equilibrium.
30 AP=50 PB or (AP/PB)=5/3
A(2,3), P(x,y), B(12,7).
30AP = 50BP,
AP = (5/3)BP,
x - 2 = 5/3(12-x),
Multiply both sides by 3:
3x - 6 = 5(12-x),
3x - 6 = 60 - 5x,
3x + 5x = 60 + 6,
8x = 66,
X = 8 2/8 = 8 1/4.
y- 3 = 5/3(7-y),
Multiply both sides by 3:
3y - 9 = 5(7-y),
3y - 9 = 35 - 5y,
3y + 5y = 35 + 9 ,
8y = 44,
Y = 5 4/8 = 5 1/2.
Solution set: P(8 1/4,5 1/2).
Well, if we're going to find a point P that can serve as the fulcrum, let's put on our balancing act hats, shall we?
We need to find the coordinates of point P that divides the distance between A and B in the ratio of 5:3.
So, here's what we'll do. We'll find the distance between A and B, and then we'll divide that distance accordingly.
The distance formula tells us that the distance between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
So, let's plug in the coordinates of A and B:
d = sqrt((12 - 2)^2 + (7 - 3)^2)
d = sqrt(10^2 + 4^2)
d = sqrt(100 + 16)
d = sqrt(116)
d ≈ 10.77
Now, we want to divide this distance in the ratio of 5:3. Let's call the distance from A to P as x, and the distance from P to B as y.
We know that x + y = 10.77 (since the sum of the two distances should be equal to the total distance).
We also know that x/y = 5/3.
Using these two equations, we can solve for x and y.
Let's multiply the second equation by 3y:
3x = 5y
Now, we can substitute this value of 3x in the first equation:
3x + x = 10.77
4x = 10.77
x = 10.77/4
x ≈ 2.69
Since x + y = 10.77, we can calculate y:
y = 10.77 - x
y ≈ 8.08
So, the coordinates of point P, which can serve as the fulcrum, are approximately (2.69, 8.08).
Now, I must say, those children may balance, but my jokes? Well, they're always a bit off-balance!
To find the point P between A(2,3) and B(12,7), which could be used as the fulcrum of a teeter board putting the two children in equilibrium, we need to find the coordinates of point P.
Let's denote the coordinates of point P as (x, y).
We know that the ratio of the distances AP and PB is equal to the ratio of the weights of the children. In this case, the weight of the 30-lb child is 30 and the weight of the 50-lb child is 50.
So, we have the equation:
AP/PB = 5/3
The distance between two points A(x₁, y₁) and B(x₂, y₂) can be calculated using the distance formula:
Distance AB = √[(x₂ - x₁)² + (y₂ - y₁)²]
Using this formula, we can calculate the distances AP and PB.
AP = √[(x - 2)² + (y - 3)²]
PB = √[(x - 12)² + (y - 7)²]
Now, we can substitute these values into the equation for the ratio and solve for x and y.
(√[(x - 2)² + (y - 3)²]) / (√[(x - 12)² + (y - 7)²]) = 5/3
Squaring both sides of the equation to eliminate the square roots:
[(x - 2)² + (y - 3)²] / [(x - 12)² + (y - 7)²] = (5/3)²
[(x - 2)² + (y - 3)²] / [(x - 12)² + (y - 7)²] = 25/9
Now, we can cross-multiply:
9[(x - 2)² + (y - 3)²] = 25[(x - 12)² + (y - 7)²]
Expanding both sides:
9[(x² - 4x + 4) + (y² - 6y + 9)] = 25[(x² - 24x + 144) + (y² - 14y + 49)]
Simplifying:
9x² - 36x + 36 + 9y² - 54y + 81 = 25x² - 600x + 3600 + 25y² - 350y + 1225
Rearranging terms:
16x² + 348x + 16y² + 296y - 3384 = 0
Now, we can solve this quadratic equation to find the coordinates (x, y) of point P.
This quadratic equation does not factorize easily, so we can use the quadratic formula:
x = [-b ± √(b² - 4ac)] / (2a)
For our equation, a = 16, b = 348, and c = 16y² + 296y - 3384.
Substituting these values into the quadratic formula, we can solve for x.
Similarly, we can solve for y by rearranging the quadratic equation in terms of y and using the quadratic formula.
Once we find the values of x and y, we will have the coordinates of point P.
Please note that the solution for point P will be in terms of x and y, and the equation provided will help find the values of x and y, but the exact coordinates cannot be determined without plugging in specific numbers for the variables.
To find the point P between A and B that could be used as the fulcrum of a teeter board, we need to find the coordinates of point P.
Let's assume the coordinates of point P are (x, y).
Given that the weight ratio is 30:50 or 3:5, we can write the equation:
AP/PB = 5/3
The distance between two points can be found using the distance formula:
Distance between two points (x1, y1) and (x2, y2) = sqrt((x2-x1)^2 + (y2-y1)^2)
Therefore, we can calculate the distances AP and PB:
AP = sqrt((x-2)^2 + (y-3)^2)
PB = sqrt((12-x)^2 + (7-y)^2)
Substituting these distances into the equation AP/PB = 5/3, we can solve for x and y.
(sqrt((x-2)^2 + (y-3)^2)) / (sqrt((12-x)^2 + (7-y)^2)) = 5/3
Next, we square both sides of the equation to eliminate the square roots:
((x-2)^2 + (y-3)^2) / ((12-x)^2 + (7-y)^2) = (5/3)^2
Simplifying, we get:
(9(x-2)^2 + 9(y-3)^2) = (25((12-x)^2 + (7-y)^2))
Expanding and rearranging terms, we have:
9x^2 - 82x + 155 + 9y^2 - 114y + 315 = 25x^2 - 600x + 3940 + 25y^2 - 350y + 1225
Combining like terms, we obtain:
16x^2 - 518x + 347y - 2850 = 0
Solving this quadratic equation will give us the values of x and y for point P, which would put the two children in equilibrium on the teeter board.