Answer:
The point on the line [tex]y=4[/tex] that is equidistant from points (0,1) and (5,6) is (2, 4).
Step-by-step explanation:
Let be [tex]y = 4[/tex], [tex]A = (0,1)[/tex] and [tex]B = (5,6)[/tex]. 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:
[tex]r_{A/P} = r_{B/P}[/tex]
Where:
[tex]r_{A/P}[/tex] - Distance of point A with respect to P.
[tex]r_{B/P}[/tex] - Distance of point B with respect to P.
We expand this equivalence by Pythagorean Theorem:
[tex]\sqrt{(x_{A}-x_{P})^{2}+(y_{A}-y_{P})^{2}} = \sqrt{(x_{B}-x_{P})^{2}+(y_{B}-y_{P})^{2}}[/tex]
[tex](x_{A}-x_{P})^{2} + (y_{A}-y_{P})^{2} = (x_{B}-x_{P})^{2}+(y_{B}-y_{P})^{2}[/tex]
[tex]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}[/tex]
[tex]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}[/tex]
And we get this expression:
[tex]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[/tex]
If we know that [tex]x_{A} = 0[/tex], [tex]y_{A} = 1[/tex], [tex]y_{P} = 4[/tex], [tex]x_{B} = 5[/tex] and [tex]y_{B} = 6[/tex], the expression is reduced to this:
[tex]0^{2}+1^{2}-5^{2}-6^{2}-2\cdot (0-5)\cdot x_{P} -2\cdot (1-6)\cdot (4) =0[/tex]
[tex]10\cdot x_{P}-20=0[/tex]
The remaining component of the point within the line is:
[tex]x_{P} = 2[/tex]
The point on the line [tex]y=4[/tex] that is equidistant from points (0,1) and (5,6) is (2, 4).
