Question

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

Answer 1

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

& \Rightarrow {{x}_{1}}=4,\ {{y}_{1}}=-6 \\\ & \Rightarrow {{x}_{2}}=-2,\ {{y}_{2}}=8 \\\ \end{aligned}$$ Let D be the distance between the points By substituting the values in the above formula, we will get, $$\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.