Question

How do you find the value of a given the points (4,-1), (a,5) with a distance of `10`?

Answer 1

`color(blue)(a = 12, -4`

Explanation:

Distance between two points is given by the formula :

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

`"Given " (x_1, y_1) = (4,-1), (x_2,y_2) = (a,5), d = 10, " To find " a`

`10 = sqrt ((a - 4)^2 + (5+1)^2)`

`(a-4)^2 + 6^2 = 10^2, " squaring both sides"`

`(a - 4)^2 = 10^2 - 6^2 = 64 = 8^2`

`a - 4 = +- 8 , " taking root on both sides"`

`color(brown)(a =) +-8 + 4 = color(brown)(12, -4`

Answer 2

`a=12, a=-4`

Explanation:

First, let's take a look at the distance formula where `d` is the distance

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

Now you choose which point is point 2 (includes `y_2` and `x_2`) and which point is point 1 (includes `y_1` and `x_1`)

Point 2: `(a,5)->y_2 = 5` and `x_2 =a`
Point 1: `(4,-1)->y_1 = -1` and `x_1 = 4`

Now plug these values into the equation

`d=sqrt((a-4)^2+(5-(-1))^2)`

We also know that the distance `d` is 10 so we can plug that in

`10=sqrt((a-4)^2+(5-(-1))^2)`

Simplify

`10=sqrt((a-4)^2+(5-(-1))^2)`

`10=sqrt((a-4)^2+(5+1)^2)`

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

`10=sqrt((a-4)^2+36)`

Now square both sides to get rid of the radical

`10^2=(sqrt((a-4)^2+36))^2`

`100=(a-4)^2+36`

FOIL the expression `(a-4)^2`

`(a-4)^2`

`(a-4)(a-4)`

`a^2+a(-4)+a(-4)+(-4)(-4)`

`a^2-4a-4a+16`

`a^2-8a+16`

Now plug this expression back in for `(a-4)^2`

`100=(a-4)^2+36`

`100=a^2-8a+16+36`

`100=a^2-8a+52`

Subtract 100 from both sides

`a^2-8a+52-100=100-100`

`a^2-8a-48=0`

Now factor this expression to find the roots/zeros

`a^2-8a-48=0`

`(a-12)(a+4)=0`

So the roots are

`a=12, a=-4`