A point (x,y) is a distance of 6 units from the x-axis. It is a distance of 5 units from the point (8,3). It is a distance √(n) from the origin. Given that x<8, what is n?

Question

A point (x,y) is a distance of 6 units from the x-axis. It is a distance of 5 units from the point (8,3). It is a distance
`√(n)` from the origin. Given that x<8, what is n?

Answer 1

Answer: n = 52

Explanation:

when we have two vectors (x,y) and (a,b) the distance between the vectors is:

D = √( (x - a)^2 + (y - b)^2)

now, we know that:

1) the distance between (x, y ) and the x-axis is 6 units.

The nearest point to (x, y) in the x-axis is the point (x, 0) so we have:

D = 6 = √( (x - x)^2 + (y - 0)^2) = √y^2

so y can be 6 or -6.

So we know that y = 6, and now we can write our point as (x, +-6)

2) The distance between our point and (8, 3) is 5 units:

D = √( (x - 8)^2 + (y - 3)^2) = 5.

And we know that the distance from the origin, (n, n) is:

D = √n = √(x^2 + y^2}

n = x^2 + y^2

Now, we should start with:

√( (x - 8)^2 + (y - 3)^2) = 5

first suppose that y = -6, then:

√( (x - 8)^2 + (-6 - 3)^2) = √( (x - 8)^2 + (-9)^2) = 5.

√( (x - 8)^2 + 81) = 5.

Then we must have that:

and we know that √25 = 5

so (x-8)^2 + 81 = 25

this can never happen, so we can discard y = -6

Now the second case, if y = 6,

√( (x - 8)^2 + (6 - 3)^2) = 5.

√( (x - 8)^2 + (3)^2) = 5.

√( (x - 8)^2 + 9) = 5.

here:

(x - 8)^2 + 9 = 25

(x - 8)^2 = 16

(x - 8) = √16 = +-4

So again we have two cases:

if x - 8 = 4, then:

x = 4 + 8 = 12

but we must have x < 8, so this can be discarded.

now, if x - 8 = -4 then:

x = -4 + 8 = 4, this is an acceptable answer, then our point is (4, 6)

And we have:

n = 4^2 + 6^2 = 16 + 36 = 52