Question

Find all the following value for x for which distance between point A(x,-1) and B(5,3) is 5units

Answer 1

Explanation:

The formula to find the distance between 2 points in the coordinate plane is

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

We have our distance, and we also have all the coordinates but the first x. Fillling in with what we have gives us this:

`5=√((5-x_1)^2+(3-(-1))^2)`

which simplifies to

`5=√((5-x_1)^2+(4)^2)` . Expanding that binomial gives us:

`5=√(25-10x+x^2+16)` . Combining like terms gives us:

`5=√(x^2-10x+41)` which is the same thing as above, only in standard form for polynomials. Now we need to get that x out from under that square root sign. We do that by squaring both sides to get:

`25=x^2-10x+41` . Now we have to factor to solve for x. We'll put everything on one side of the equals sign, set the polynomial equal to 0, then factor.

`0=x^2-10x+16` is our polynomial now. a = 1, b = -10, c = 16. The product ac is 1 * 16 which is 16. Some combination of the factors of 16 will result in a -10. So we need the factors of 16.

16: {1, 16}, {2, 8}, {4, 4}

The only combination of those factors that will result in a -10 is the second pair, {2, 8}. If we add 2 and 8 we get 10, but in order for our 10 to be negative, both 2 and 8 have to be negative. So we rewrite the polynomial in terms of -2 and -8:

`0=x^2-8x-2x+16`

Now we can factor by grouping. Group the first 2 terms together and the second 2 terms together without moving any of their positions:

`0=(x^2-8x)-(2x+16)`

From each set of parenthesis we will now factor out what's common. x is common in the first set of ( ), and 2 is common in the second set of ( ):

`0=x(x-8)-2(x-8)`

What's common now is the binomial (x - 8). So we'll factor that out now:

`0=(x-8)(x-2)`

By the Zero Product Property, either

x - 8 = 0 or x - 2 = 0.

If x - 8 = 0, then x = 8. If x - 2 = 0, then x = 2.

It looks like we have 2 solutions. Let's try them both and see if, when we stick an 8 and then a 2 into our distance formula, the distance is 5:

`d=√((5-8)^2+(4^2))` is

`d=√((-3)^2+(4)^2)` is

`d=√(9+16)` is

`d=√(25)` which does in fact equal 5. Now let's try the 2:

`d=√((5-2)^2+(4)^2)` which is

`d=√((3)^2+(4)^2)` is

`d=√(9+16)` is

`d=√(25)` which also comes out to equal 5.

So the 2 values of x which will work here are 2 and 8.

Answer 2

x = 2, x = 8

Explanation:

Calculate the distance using the distance formula and equate to 5

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = A (x, - 1) and (x₂, y₂ ) = B(5, 3)

d =
`√((5-x)^2+(3+1)^2)`

=
`√((5-x)^2+4^2)`

=
`√((5-x)^2+16)` , thus

`√((5-x)^2+16)` = 5 ( square both sides )

(5 - x)² + 16 = 25 ( subtract 16 from both sides )

(5 - x)² = 9 ( take the square root of both sides )

5 - x = ±
`√(9)` = ± 3 ( subtract 5 from both sides )

- x = - 5 ± 3 , thus

- x = - 5 + 3 = - 2 ( multiply both sides by - 1 )

x = 2

or

x = - 5 - 3 = - 8 ( multiply both sides by - 1 )

x = 8