Question

find the locus of a point equidistance from a point(2,4) and the y axis

Answer 1

Find the locus of a point equidistance from a point(2,4) and the y axis.

Answer:

The point situated equidistance from a point $\mathsf{P}(\mathsf{x},\mathsf{y})$ and the y axis located in the line connected a point P and y axis by the right angel (the nearest distance), according this line is parallel to x-axis and all points of it have equal coordinate Y. A distance from M to P equal to distance from M and y axis according this the X coordinate of M is as $\frac{1}{2}\mathbf{X}$ of coordinate P

The point situated equidistance from a point $\mathsf{P}$ have coordinates:

$M(x_{1},y_{1})$

$y_{1} = y$

$x_{1} = \frac{1}{2} x;$

Answer: point- (1, 4)