Question

How do you find the distance and midpoint between the two points. \[\left( 4,-6 \right)\] \[\left( -2,8 \right)\]?

Answer 1

Hint: From the given question we have to find the distance and midpoint between the points \[\left( 4,-6 \right)\] and \[\left( -2,8 \right)\]. we know that formula of distance of two points and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] and the midpoint between the two points is literally just the average between the x values and the average between the y values i.e., \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]. By these two formulas we will get the required solutions.

Complete step by step solution:
From the question we have two points they are,
 \[\Rightarrow \left( 4,-6 \right),\left( -2,8 \right)\]
Firstly, we have to find the distance between these points.
 We know that formula for the distance between the points and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is
\[\Rightarrow \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
Therefore, here by comparing
\[\begin{align}
  & \Rightarrow {{x}_{1}}=4,\ {{y}_{1}}=-6 \\
 & \Rightarrow {{x}_{2}}=-2,\ {{y}_{2}}=8 \\
\end{align}\]
Let D be the distance between the points
By substituting the values in the above formula, we will get,
\[\Rightarrow D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -2-4 \right)}^{2}}+{{\left( 8-\left( -6 \right) \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( 14 \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{36+196}\]
\[\Rightarrow D=\sqrt{232}\]
\[\Rightarrow D=\sqrt{58\times 4}\]
\[\Rightarrow D=2\sqrt{58}\]
Therefore, the distance between the two points is \[D=2\sqrt{58}\]
Now, we have to find the midpoint of the above two points.
Already we discussed that the midpoint between the two points is literally just the average between the x values and the average between the y values i.e., \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\].
Let M is the midpoint of the above two points.
By substituting the above values in the above formula, we will get,
\[\Rightarrow M=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
\[\Rightarrow M=\left( \dfrac{4+\left( -2 \right)}{2},\dfrac{-6+8}{2} \right)\]
\[\Rightarrow M=\left( \dfrac{2}{2},\dfrac{2}{2} \right)\]
\[\Rightarrow M=\left( 1,1 \right)\]
Therefore, the midpoint of the above two points is \[ M=\left( 1,1 \right)\]

Note:
Students should know the basic formulas of the coordinate system. To find the distance we just apply Pythagoras. Think of it this way:
The difference between the x points causes a straight horizontal line, the difference between the y points causes a straight vertical line, so the distance between the two points is the hypotenuse i.e., \[\Rightarrow D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]. Students should be careful while doing calculations and they should notice the signs of numbers.