Question

In a coordinate plane, how many points are both 5 units from the origin and 2 units from the X-axis?

Answer 1

we first try to form a point in the coordinate plane in the form of $A\equiv \left( x,y \right)$. We form the given condition into mathematical form and solve them. We solve the equation to find the number of possible points.

Complete step by step answer:
Any point on the coordinate plane can be expressed in the form of $A\equiv \left( x,y \right)$.
The individual values of the coordinates indicate its distance from the axes. For $A\equiv \left( x,y \right)$, the distance of the point from the X axis and y axis is $\left| y \right|$ and $\left| x \right|$ units respectively.
The distance of any point $A\equiv \left( x,y \right)$ from the origin $O\equiv \left( 0,0 \right)$ is $\left| A \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ units.
For our given problem, the point has to be 5 units from the origin and 2 units from the X-axis.
We can form the mathematical form of the given conditions where $\sqrt{{{x}^{2}}+{{y}^{2}}}=5$ and $\left| y \right|=2$.
We have two equations and two unknowns to solve.
From $\left| y \right|=2$, we get two solutions for $y$ where $y=\pm 2$.
We can take square of the equation $\left| y \right|=2$ and get

& {{\left| y \right|}^{2}}={{2}^{2}} \\\ & \Rightarrow {{y}^{2}}=4 \\\ \end{aligned}$$ Putting the value in the equation of $\sqrt{{{x}^{2}}+{{y}^{2}}}=5$, we get $\sqrt{{{x}^{2}}+4}=5$. Now taking square we get $\begin{aligned} & {{x}^{2}}+4={{5}^{2}}=25 \\\ & \Rightarrow {{x}^{2}}=25-4=21 \\\ & \Rightarrow x=\pm \sqrt{21} \\\ \end{aligned}$ Therefore, the points can be in the form of $\left( \pm \sqrt{21},\pm 2 \right)$. There are four possible points which are $\left( \sqrt{21},2 \right)$, $\left( \sqrt{21},-2 \right)$, $\left( -\sqrt{21},2 \right)$, $\left( -\sqrt{21},-2 \right)$. **Note:** We have to remember that although the value of a point in the X axis is $x$ but its perpendicular distance from the X-axis is $y$ units. We took modulus values to define both positive and negative sides.