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

To find coordinates equidistant from point F=(2, 3) and the y-axis, we equate the square of the distance from a point P=(x, y) to F with the square of the distance from P to the y-axis. Example points include (2, 3), (1, 3+√2), and (1, 3-√2), and the equation representing such points is (x - 2)² + (y - 3)² = x².

Step-by-step explanation:

To find coordinates that are equidistant from the fixed point F and the y-axis, we need to use the definition of distance in the Cartesian coordinate system. Let's consider the point F=(2, 3) and a generic point P=(x, y) that is equidistant from F and the y-axis. The distance from P to F is the same as the distance from P to the y-axis because P should be equidistant from both.

To calculate the distance from P to F, we use the distance formula, which is the square root of the sum of the squared differences in x and y coordinates. However, since we are looking for a point equidistant from F and the y-axis, we can equate the two distances without the square root to make the calculations simpler. The distance PF is given by:

√((x - 2)² + (y - 3)²)

Since the distance to the y-axis for any point P is just its x-coordinate's absolute value, because the y-axis is vertical line at x=0, we have:

|x|.

Equating the squares of these distances:

(x - 2)² + (y - 3)² = x²

Examples of such points could be (2, 3), (1, 3+√2), and (1, 3-√2), by solving the above equation for various values of y while keeping x constant. The equation to state that P=(x, y) is equidistant from F and the y-axis is:

(x - 2)² + (y - 3)² = x²

Answer 2

The equation that says P is equidistant from F and the y-axis is
`P(x,y) =\left(1,(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) = (1)/(2)\cdot F(x,y)+(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,(3)/(2) \right)+\left(0, (y')/(2)\right)`

`P(x,y) =\left(1,(3+y')/(2) \right)`

The equation that says P is equidistant from F and the y-axis is
`P(x,y) =\left(1,(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,(3+y_(1)')/(2) \right)`

`P_(1)(x,y) = (1,0)`

`P_(2)(x,y) = \left(1,(3+y_(2)')/(2) \right)`

`P_(2)(x,y) = \left(1,(3)/(2) \right)`

`P_(3)(x,y) = \left(1,(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.