Respuesta :
Answer:
The equation that says P is equidistant from F and the y-axis is [tex]P(x,y) =\left(1,\frac{3+y'}{2} \right)[/tex].
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.
Step-by-step explanation:
Let [tex]F(x,y) = (2,3)[/tex] and [tex]R(x,y) =(0, y')[/tex], where [tex]P(x,y)[/tex] is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:
[tex]F(x,y)-P(x,y) = P(x,y)-R(x,y)[/tex]
[tex]2\cdot P(x,y) = F(x,y)+R(x,y)[/tex]
[tex]P(x,y) = \frac{1}{2}\cdot F(x,y)+\frac{1}{2} \cdot R(x,y)[/tex] (1)
If we know that [tex]F(x,y) = (2,3)[/tex] and [tex]R(x,y) = (0,y')[/tex], then the resulting vectorial equation is:
[tex]P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)[/tex]
[tex]P(x,y) =\left(1,\frac{3+y'}{2} \right)[/tex]
The equation that says P is equidistant from F and the y-axis is [tex]P(x,y) =\left(1,\frac{3+y'}{2} \right)[/tex].
If we know that [tex]y_{1}' = -3[/tex], [tex]y_{2}' = 0[/tex] and [tex]y_{3}' = 3[/tex], then the coordinates for three points that are equidistant from F and the y-axis:
[tex]P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)[/tex]
[tex]P_{1}(x,y) = (1,0)[/tex]
[tex]P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)[/tex]
[tex]P_{2}(x,y) = \left(1,\frac{3}{2} \right)[/tex]
[tex]P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)[/tex]
[tex]P_{3}(x,y) = \left(1,6 \right)[/tex]
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.