Question

How do you find the possible values for a if the points (a,6), (-6,2) has a distance of `4sqrt10`?

Answer 1

See a solution process below:

Explanation:

The formula for calculating the distance between two points is:

`d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)`

Substituting the values for the distance and from the points in the problem gives:

`4sqrt(10) = sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2)`

We can now solve for `a`:

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

`(4sqrt(10))^2 = (sqrt((color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2))^2`

`16 * 10 = (color(red)(-6) - color(blue)(a))^2 + (color(red)(2) - color(blue)(6))^2`

`160 = (color(red)(-6) - color(blue)(a))^2 + (-4)^2`

`160 = (color(red)(-6) - color(blue)(a))^2 + 16`

`160 = (-6)^2 + 6a + 6a + a^2 + 16`

`160 = 36 + 12a + a^2 + 16`

`160 = a^2 + 12a + 52`

`160 - color(red)(160) = a^2 + 12a + 52 - color(red)(160)`

`0 = a^2 + 12a -108`

We can factor this as:

#0 = (a + 18)(a - 6)

Now, equate each term on the right side of the equation to `0` and solve for `a`:

Solution 1)

`a + 18 = 0`

`a + 18 - color(red)(18) = 0 - color(red)(18)`

`a + 0 = -18`

`a = -18`

Solution 3)

`a - 6 = 0`

`a - 6 + color(red)(6) = 0 + color(red)(6)`

`a - 0 = 6`

`a = 6`

The solution is, a could be `-18` or `6`

Answer 2

`a=-18`
or
`a=6`

Explanation:

Distance between two points, `(x_1,y_1)` and `(x_2,y_2)`

= `sqrt((x_1-x_2)^2+(y_1-y_2)^2`

In this case
`(a,6)` and `(-6,2)`

`4sqrt(10)=sqrt([a-(-6)]^2+(6-2)^2)`
`(4sqrt(10))^2={sqrt((a+6)^2+(4)^2)}^2`
`16*10=(a+6)^2+16`
`160=a^2+6x+6x+36+16`
`a^2+12x-108=0`

Carry out factorisation
`a^2+12x-108=0`
`(a+18)(a-6)=0`

Divide RHS `(0)` with `(a+18)` and `(a-6)`
You will have
`(a+18)=0` and `(a-6)=0` which give you
`a=-18` or `a=6` respectively