
What is the distance between the Point A $\left( 2,-3 \right)$ and the Point B $\left( 2,4 \right)$ ?
Answer
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Hint: Here, we need to find out the distance between the two give points. First, we write the formula for the distance between two points. After that, we substitute the given points in the formula to get the distance.
Complete step-by-step answer:
Coordinate systems are of various types. The major types are Cartesian coordinate system, Cylindrical coordinate system and the Spherical coordinate system. Any point in the Cartesian coordinate system is represented as $\left( x,y,z \right)$ in three dimension and $\left( x,y \right)$ in two dimensions. The first part is called the abscissa or the x-coordinate and the second part is called the ordinate or the y-coordinate.
Let there be two points on the x-y plane, which are $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ . The distance between the two points is the length of the line segment joining them, which is AB. Now, if we observe closely, we find that the triangle $\Delta ABC$ is a right-angled triangle with the right angle at C. The length BC will be ${{x}_{2}}-{{x}_{1}}$ and that of AC is ${{y}_{2}}-{{y}_{1}}$ . The length AB will then be, by Pythagoras theorem,
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
We are given that the two points are $\left( 2,-3 \right)$ and $\left( 2,4 \right)$ . Substituting them in the above formula, we get,
$\begin{align}
& \Rightarrow AB=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 4-\left( -3 \right) \right)}^{2}}} \\
& \Rightarrow AB=7units \\
\end{align}$
Therefore, we can conclude that the distance between the two points is $7units$ .
Note: This problem can also be solved in another way. If we observe carefully, the two points $\left( 2,-3 \right)$ and $\left( 2,4 \right)$ lie on the same $x=2$ line. This means that we can simply subtract the ordinates of the two points, keeping in mind that the distance cannot be negative. The answer will be $4-\left( -3 \right)=7$ .
Complete step-by-step answer:
Coordinate systems are of various types. The major types are Cartesian coordinate system, Cylindrical coordinate system and the Spherical coordinate system. Any point in the Cartesian coordinate system is represented as $\left( x,y,z \right)$ in three dimension and $\left( x,y \right)$ in two dimensions. The first part is called the abscissa or the x-coordinate and the second part is called the ordinate or the y-coordinate.
Let there be two points on the x-y plane, which are $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ . The distance between the two points is the length of the line segment joining them, which is AB. Now, if we observe closely, we find that the triangle $\Delta ABC$ is a right-angled triangle with the right angle at C. The length BC will be ${{x}_{2}}-{{x}_{1}}$ and that of AC is ${{y}_{2}}-{{y}_{1}}$ . The length AB will then be, by Pythagoras theorem,
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
We are given that the two points are $\left( 2,-3 \right)$ and $\left( 2,4 \right)$ . Substituting them in the above formula, we get,
$\begin{align}
& \Rightarrow AB=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 4-\left( -3 \right) \right)}^{2}}} \\
& \Rightarrow AB=7units \\
\end{align}$
Therefore, we can conclude that the distance between the two points is $7units$ .
Note: This problem can also be solved in another way. If we observe carefully, the two points $\left( 2,-3 \right)$ and $\left( 2,4 \right)$ lie on the same $x=2$ line. This means that we can simply subtract the ordinates of the two points, keeping in mind that the distance cannot be negative. The answer will be $4-\left( -3 \right)=7$ .
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