Question

How do you find a possible value for a if the points (-2,a), (6,1)has a distance of `d=4sqrt5`?

Answer 1

`a=-3" or "a=5`

Explanation:

`"using the "color(blue)"distance formula"`

`•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y_1)=(6,1)" and "(x_2,y_2)=(-2,a)`

`d=sqrt((-2-6)^2+(a-1)^2)=4sqrt5`

`rArrsqrt(64+(a-1)^2)=4sqrt5`

`color(blue)"square both sides"`

`rArr64+(a-1)^2=80`

`"subtract 64 from both sides"`

`(a-1)^2=16`

`color(blue)"take the square root of both sides"`

`rArra-1=+-sqrt16larrcolor(blue)"note plus or minus"`

`rArra=1+-4`

`rArra=1-4=-3" or "a=1+4=5`

Answer 2

Possible values of `a` are `-3 and 5`.

Explanation:

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

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

Distance between two points `(-2,a) and (6,1)` is

`D= sqrt((-2-6)^2+(a-1)^2)= 4sqrt5` , Squaring both

sides we get , `64+ a^2-2a+1=80` or

`a^2-2a+1+64-80=0` or

`a^2-2a-15=0 or a^2-5a+3a-15=0` or

`a(a-5)+3(a-5)=0 :. (a+3)(a-5)=0`

`:. a =-3, a = 5`

Possible values of `a` are `-3 and 5` [Ans]