Question

How do you find the possible values for a if the points (5,8), (a,2) has a distance of `3sqrt5`?

Answer 1

Use the distance formula and solve for `a`:
`{a:a=2,8}`

Explanation:

The distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is given as `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`.

For this problem, it is given that `(x_1,y_1)=(5,8)`, `(x_2,y_2)=(a,2)`, and `d=3sqrt5`. We need to solve for `a` by plugging these values into the distance formula.
`3sqrt5=sqrt((a-5)^2+(2-8)^2)`.

To solve for `a`, begin by squaring both sides of the equation and simplify a bit.
`(3sqrt5)^color(green)2=sqrt((a-5)^2+(2-8)^2)^color(green)2`
`3^2sqrt5^2=sqrt((a-5)^2+(2-8)^2)^2`
`9*5=(a-5)^2+(2-8)^2`
`45=(a-5)^2+("-"6)^2`
`45=a^2-10a+25+36`
`45=a^2-10a+61`

Now subtract 45 from both sides to get an equation in quadratic form.
`45color(green)(-45)=a^2-10a+61color(green)(-45)`
`0=a^2-10a+16`

Finally, solve for `a` using either the quadratic formula or the factoring method. I'll be factoring.
`0=a^2-2a-8a+16`
`0=a(a-2)-8(a-2)`
`0=(a-8)(a-2)`
`a=8or2`

Therefore the possible values for `a` are defined by the set `{a:a=2,8}`