Can you please explain how I would solve these questions?
1) Which point is 2 units from the line 12x-9y-5=0?
a)(8,3)
b)(7/6, -7/3)
c)(14,6)
d)(1,1)
2.) Which of the following sets of vectors spans R2:
a) {(1,1),(-2,-2)}
b) {(1,1),(1,2)}
c) {(1,2),(1/2,2)}
d) {(-1,1),(1,-1)}
For any point (a,b)
|12a - 9b - 5| /√(144+81) = 2
12a - 9b - 5 = 30
12a - 9b = 35
Now test which ordered pair satisfies this linear equation.
Sure! Let's break down each question and explain how to solve them.
1) Which point is 2 units from the line 12x-9y-5=0?
To solve this question, we can use the distance formula between a point and a line. The distance formula between a point (x1, y1) and a line Ax + By + C = 0 is given by:
Distance = |Ax1 + By1 + C| / √(A^2 + B^2)
Here, A = 12, B = -9, C = -5. We need to find which point among the options is 2 units away from this line.
For each point (x, y) in the options, substitute the values of x and y into the distance formula and check if the result is equal to 2.
Let's calculate the distances:
a) (8,3):
Distance = |12(8) - 9(3) - 5| / √(12^2 + (-9)^2)
Distance = |96 - 27 - 5| / √(144 + 81)
Distance = |64| / √(225)
Distance = 64 / 15
The distance is not equal to 2, so point (8,3) is not the answer.
b) (7/6, -7/3):
Distance = |12(7/6) - 9(-7/3) - 5| / √(12^2 + (-9)^2)
Distance = |14 - 21/2 - 5| / √(144 + 81)
Distance = |28 - 21 - 10| / √(225)
Distance = 3 / 15
The distance is equal to 1/5, so point (7/6, -7/3) is not the answer.
c) (14,6):
Distance = |12(14) - 9(6) - 5| / √(12^2 + (-9)^2)
Distance = |168 - 54 - 5| / √(144 + 81)
Distance = |109| / √(225)
Distance = 109 / 15
The distance is not equal to 2, so point (14,6) is not the answer.
d) (1,1):
Distance = |12(1) - 9(1) - 5| / √(12^2 + (-9)^2)
Distance = |12 - 9 - 5| / √(144 + 81)
Distance = |-2| / √(225)
Distance = 2 / 15
The distance is not equal to 2, so point (1,1) is not the answer.
None of the given points (a, b, c, d) is 2 units away from the line 12x-9y-5=0. Therefore, the answer is none of the above.
2) Which of the following sets of vectors spans R2?
To determine which set of vectors spans R2, we need to check if any linear combination of vectors from the set can produce any vector in R2.
a) {(1,1),(-2,-2)}:
We can check if any vector (x, y) in R2 can be obtained as a linear combination of (1,1) and (-2,-2). Solve the system of equations:
x(1,1) + y(-2,-2) = (x, y)
This system can be written as:
x - 2y = x
x - 2y = y
In both cases, we can see that no matter what values of x and y are chosen, the system of equations cannot be satisfied. Therefore, the set {(1,1),(-2,-2)} does not span R2.
b) {(1,1),(1,2)}:
Using the same approach as before, we need to check if any vector (x, y) in R2 can be obtained as a linear combination of (1,1) and (1,2). Solve:
x(1,1) + y(1,2) = (x, y)
This system can be written as:
x + y = x
x + 2y = y
Again, no matter what values of x and y are chosen, the system of equations cannot be satisfied. Therefore, the set {(1,1),(1,2)} does not span R2.
c) {(1,2),(1/2,2)}:
We can check if any vector (x, y) in R2 can be obtained as a linear combination of (1,2) and (1/2,2). Solve:
x(1,2) + y(1/2,2) = (x, y)
This system can be written as:
x + (1/2)y = x
2x + 2y = y
Once again, no matter what values of x and y are chosen, the system of equations cannot be satisfied. Therefore, the set {(1,2),(1/2,2)} does not span R2.
d) {(-1,1),(1,-1)}:
Check if any vector (x, y) in R2 can be obtained as a linear combination of (-1,1) and (1,-1). Solve:
x(-1,1) + y(1,-1) = (x, y)
This system can be written as:
-x + y = x
x - y = y
Once again, the system of equations cannot be satisfied for any values of x and y. Therefore, the set {(-1,1),(1,-1)} does not span R2.
Based on the analysis above, none of the given sets of vectors (a, b, c, d) spans R2.