Question

Find a point on the line y = 4 that is equidistant from the two points (0,1) and (5,6)

Answer 1

The point on the line
`y=4` that is equidistant from points (0,1) and (5,6) is (2, 4).

Explanation:

Let be
`y = 4`,
`A = (0,1)` and
`B = (5,6)`. As we know that given function is a horizontal line, the condition of equidistance between that a point within that line and both points must be:

`r_(A/P) = r_(B/P)`

Where:

`r_(A/P)` - Distance of point A with respect to P.

`r_(B/P)` - Distance of point B with respect to P.

We expand this equivalence by Pythagorean Theorem:

`\sqrt{(x_(A)-x_(P))^(2)+(y_(A)-y_(P))^(2)} = \sqrt{(x_(B)-x_(P))^(2)+(y_(B)-y_(P))^(2)}`

`(x_(A)-x_(P))^(2) + (y_(A)-y_(P))^(2) = (x_(B)-x_(P))^(2)+(y_(B)-y_(P))^(2)`

`x_(A)^(2)-2\cdot x_(A)\cdot x_(P)+x_(P)^(2) + y_(A)^(2)-2\cdot y_(A)\cdot y_(P)+y_(P)^(2) = x_(B)^(2)-2\cdot x_(B)\cdot x_(P)+x_(P)^(2) + y_(B)^(2)-2\cdot y_(B)\cdot y_(P)+y_(P)^(2)`

`x_(A)^(2)-2\cdot x_(A)\cdot x_(P) + y_(A)^(2)-2\cdot y_(A)\cdot y_(P) = x_(B)^(2)-2\cdot x_(B)\cdot x_(P) + y_(B)^(2)-2\cdot y_(B)\cdot y_(P)`

And we get this expression:

`x_(A)^(2)+y_(A)^(2)-x_(B)^(2)-y_(B)^(2) - 2\cdot (x_(A)-x_(B))\cdot x_(P)-2\cdot (y_(A)-y_(B))\cdot y_(P) = 0`

If we know that
`x_(A) = 0`,
`y_(A) = 1`,
`y_(P) = 4`,
`x_(B) = 5` and
`y_(B) = 6`, the expression is reduced to this:

`0^(2)+1^(2)-5^(2)-6^(2)-2\cdot (0-5)\cdot x_(P) -2\cdot (1-6)\cdot (4) =0`

`10\cdot x_(P)-20=0`

The remaining component of the point within the line is:

`x_(P) = 2`

The point on the line
`y=4` that is equidistant from points (0,1) and (5,6) is (2, 4).