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Aren't planes described by $(x,y,z)$?

I know that the point (2,0) satisfies a, and the point (1.5,4) satisfies b. But how would I turn these into equations and geometric descriptions?

I'm stuck.

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    By "plane" they mean the space of ordered pairs $(x,y)$, rather than two-dimensional planes in 3-dimensional space. – 2017-01-25
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    Why stop at dimension 3? Planes (2-planes) exist in higher dimensions as well. The answer is that you can always find a coordinate system for the larger space in which all but two coordinates are constant on the plane. So, just suppress them all and consider $\mathbb R^2$ instead. – 2017-01-25
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    "...the set of **ALL** points..." For example, the set of all points $(x,y)$ in the plane which are $1$ unit away from the origin is described by $x^2+y^2=1$. – 2017-01-25
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    @John here's a hint: how could you **check** that $(2,0)$ satisfies a)? If I gave you a random point $(x,y)$, how would you **check** that it satisfies a)? Same for b) – 2017-01-25

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By the Pythagorean theorem, the distance between two points $(x,y)$ and $(a,b)$ is given by $\sqrt{(x-a)^2+(x-b)^2}$. The condition given in a) then translates to $$\sqrt{(x-4)^2+y^2}=\sqrt{(x-1)^2+y^2}.$$ We can solve this for $y$. Take the square and simplify to obtain $$y=\pm\sqrt{4-x^2}.$ This equation characterizes $S$. It is a circle around the origin with radius $2$.

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Hint: When trying to form equations from descriptions, "is" usually corresponds to an equal sign. "The distance between a point $(a,b)$ and a point $(c,d)$" is the quantity $\sqrt{(a-c)^2+(b-d)^2}$. Do you know how to find the distance between a point and a vertical line?

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    I think you misunderstand. For (b), you want the distance from a generic point $(a,b)$ to $(2,4)$ to equal the distance from the same point $(a,b)$ to the line "$x=-1$". – 2017-01-25
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Hint: Using the distance formula, a point $(x,y)$ will satisfy a) exactly when $$ \sqrt{(x-4)^2 + y^2} = 2\sqrt{(x-1)^2 + y^2} = 1 $$ and it will satisfy b) if $$ |x+1| = \sqrt{(x-2)^2 + (y-4)^2} $$ However, both of these equations can be simplified.

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    @John the distance between the point $(x,y)$ and the line $x = -1$ is the same as the distance between the point $(x,y)$ and $(-1,y)$, the closest point on the line to $(x,y)$. – 2017-01-25