Question

Let F = (2, 3). Find coordinates for three points that are equidistant from F and the y-axis. Write an equation that says P = (x, y) is equidistant from F and the y-axis

Answer 1

Answer:

The equation that says P is equidistant from F and the y-axis is $P(x,y) =\left(1,\frac{3+y'}{2} \right)$.

(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 $F(x,y) = (2,3)$ and $R(x,y) =(0, y')$, where $P(x,y)$ 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:

$F(x,y)-P(x,y) = P(x,y)-R(x,y)$

$2\cdot P(x,y) = F(x,y)+R(x,y)$

$P(x,y) = \frac{1}{2}\cdot F(x,y)+\frac{1}{2} \cdot R(x,y)$ (1)

If we know that $F(x,y) = (2,3)$ and $R(x,y) = (0,y')$, then the resulting vectorial equation is:

$P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)$

$P(x,y) =\left(1,\frac{3+y'}{2} \right)$

The equation that says P is equidistant from F and the y-axis is $P(x,y) =\left(1,\frac{3+y'}{2} \right)$.

If we know that $y_{1}' = -3$, $y_{2}' = 0$ and $y_{3}' = 3$, then the coordinates for three points that are equidistant from F and the y-axis:

$P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)$

$P_{1}(x,y) = (1,0)$

$P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)$

$P_{2}(x,y) = \left(1,\frac{3}{2} \right)$

$P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)$

$P_{3}(x,y) = \left(1,6 \right)$

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.