Explain this to me look below.Its Algebra?

Question

Find the distance between each pair of points.Where appropriate find an approximation to three decimal places.It says leave answers in (Radical Form)

1) (1,6) and (5,9)

2) (0,-7) and (7,2)

3) (-4,4) and (6,-6)

4) (-4,-2) and (-7,-11)

Answer 1

If (x1, y1) and (x2, y2) are your two points, then the distance between them is
√[ (x1 - x2)² + (y1 - y2)² ]

So using the first example:
√[ (5 - 1)^2 + (9 - 6)^2 ] = √[ 16 + 9 ] = √25 = 5

Answer 2

If the points are (a,b) and (c,d), the distance between them is:
sqrt[ (d - b)^2 + (c - a)^2 ].

The instructions to evaluate to 3 decimal places and to leave in radical form seem to contradict each other. I have left them in radical form where they do not come out exactly. If you want 3 dec. places, use a calculator to find the square root.

1.
Using the above formula, the distance is:
sqrt[ (9 - 6)^2 + (5 - 1)^2 ]
= sqrt[ 3^2 + 4^2 ]
= sqrt[ 9 + 16 ]
= sqrt(25)
= 5.

Similarly for the others.

2.
sqrt[ (2 - (-7))^2 + (7 - 0)^2 ]
= sqrt( 81 + 49)
= sqrt( 130 ).

3.
sqrt(100 + 100)
= sqrt(200)
= 10sqrt(2).

4.
sqrt(81 + 9)
= sqrt(90)
= 3sqrt(10).

Answer 3

This is really just a case of applying the distance formula: d((x₁, y₁), (x₂, y₂)) = √((x₂-x₁)² + (y₂-y₁)²). So for instance, the distance from (1, 6) to (5, 9) is simply √((5-1)²+(9-6)²) = √(4²+3²) = √(16+9) = √25 = 5. The other problems are solved in a similar manner.

The reason why the distance formula takes this form has to do with the Pythagorean theorem -- basically, you can form a right triangle where one leg goes horizontally from x₁ to x₂ (and thus has length |x₂-x₁|), and the other leg goes vertically from y₁ to y₂ (and thus has length |y₂-y₁|). The straight line connecting those points will then be the hypotenuse of this triangle, and the square of its length will be the sum of the squares of the other two sides. Thus, d((x₁, y₁), (x₂, y₂))² = (x₂-x₁)² + (y₂-y₁)² and so d((x₁, y₁), (x₂, y₂)) = √((x₂-x₁)² + (y₂-y₁)²).

Answer 4

Use Pythagoras' Formula
C^2 = A^2 + B^2
Where C is the Distance
A is the X difference
B is the Y difference
Or, restated as the Difference formula...
D = (X^2 + Y^2)^(1/2)
For example, given the points (x1,y1) and (x2,y2)
D = ((x2-x1)^2 + (y2-y1)^2)^(1/2)

1) D = ((5-1)^2 + (9-6)^2)^(1/2)
= ((4)^2 + (3)^2)^(1/2)
= (16 + 9)^(1/2)
= (25)^(1/2)
= 5
2) Gives, by a similar process
D = (130)^(1/2), which is leaving it in the Radical form

I leave the rest to you. Happy hunting. Use your Calculator.

Answer 5

For all of these problems to you can use the distance formula:

d = √(x2 - x1)^2 + (y2 - y1)^2

You can name either point as being (x1, y1) and (x2, y2). The key is that you remain consistent with the order. So for the first problem:

1) (1,6) and (5,9)

d = √(5 -1)^2 + (9 - 6)^2
d = √(4)^2 + (3)^2
d = √16 + 9
d = √25
d = 5

You can use the same formula and method for all of the problems. If you try the rest, it'll give you good practice.

Answer 6

The distance formula is really the Pythagorean theorem.

Use the two sets of points given (1,6) (5,9) as the ends of the hypotenuse of a right triangle.
The bottom leg will run from (1,6) to (5,6), a distance of 4
The remaining leg will run from (5,6) to (5,9) a distance of 3

3^2 + 4^2 = 9+16 =25. The distance from (1,6) to (5,9) is therefore the square root of 25, or 5 units.

Answer 7

You have the formula for the distance between two points.
d(P1P2) = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
The instructions say to both "find an approximation..." & "leave answers in radical form". Do they define "where appropriate"?

Answer 8

1. sqrt(4^2+3^2)=5
2. sqrt(7^2+9^2)=sqrt(130)=11.401
3. sqrt(10^2+10^2)=sqrt(200)=14.142
4. sqrt(3^2+9^2)=sqrt(90)=9.486

Answer 9

         ___________________
D = √(x₁- x₂)² + (y₁- y₂)²