How do you find the distance between the \[\left( -3,-5 \right)\]\[\left( -6,-8 \right)\]?
Answer
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Hint: From the given question we have to find the distance and midpoint between the points \[\left( -3,-5 \right)\] and \[\left( -6,-8 \right)\]. we know that formula of distance of two points and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\].By the above formulas we will get the required solution.
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
\[\Rightarrow {{d}^{2}}=\Delta {{x}^{2}}+\Delta {{y}^{2}}\]
\[\Rightarrow d=\sqrt{\Delta {{x}^{2}}+\Delta {{y}^{2}}}\]
Here d is the distance between the two points.
\[\Delta x\] is the difference between the x points that causes a straight horizontal line.
\[\Rightarrow \Delta x={{x}_{2}}-{{x}_{1}}\]
\[\Delta y\] is the difference between the x points that causes a straight horizontal line.
\[\Rightarrow \Delta y={{y}_{2}}-{{y}_{1}}\]
Complete step by step solution:
From the question we have two points they are,
\[\Rightarrow \left( -3,-5 \right),\left( -6,-8 \right)\]
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
\[\begin{align}
& \Rightarrow {{x}_{1}}=-3,\ {{y}_{1}}=-5 \\
& \Rightarrow {{x}_{2}}=-6,\ {{y}_{2}}=-8 \\
\end{align}\]
Let D be the distance between the points
By substituting the values in the above formula, we will get,
\[\Rightarrow D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -6-\left( -3 \right) \right)}^{2}}+{{\left( -8-\left( -5 \right) \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{9+9}\]
\[\Rightarrow D=\sqrt{18}\]
\[18\]can be written as product of \[9\]and \[2\]
\[\Rightarrow D=\sqrt{9.2}\]
Therefore \[9\] is the square of \[3\]
\[\Rightarrow D=3\sqrt{2}\]
Therefore, the distance between the two points is \[ D=3\sqrt{2}\]
Note:
Students should be very careful while doing the calculation like, \[ D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] here in this formula there itself negative signs are there students should substitute the exact values of the points with their signs, if we write \[3\] in place of \[{{x}_{1}}\] then answer will be changed we should have to write \[-3\] only.
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
\[\Rightarrow {{d}^{2}}=\Delta {{x}^{2}}+\Delta {{y}^{2}}\]
\[\Rightarrow d=\sqrt{\Delta {{x}^{2}}+\Delta {{y}^{2}}}\]
Here d is the distance between the two points.
\[\Delta x\] is the difference between the x points that causes a straight horizontal line.
\[\Rightarrow \Delta x={{x}_{2}}-{{x}_{1}}\]
\[\Delta y\] is the difference between the x points that causes a straight horizontal line.
\[\Rightarrow \Delta y={{y}_{2}}-{{y}_{1}}\]
Complete step by step solution:
From the question we have two points they are,
\[\Rightarrow \left( -3,-5 \right),\left( -6,-8 \right)\]
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
\[\begin{align}
& \Rightarrow {{x}_{1}}=-3,\ {{y}_{1}}=-5 \\
& \Rightarrow {{x}_{2}}=-6,\ {{y}_{2}}=-8 \\
\end{align}\]
Let D be the distance between the points
By substituting the values in the above formula, we will get,
\[\Rightarrow D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -6-\left( -3 \right) \right)}^{2}}+{{\left( -8-\left( -5 \right) \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}\]
\[\Rightarrow D=\sqrt{9+9}\]
\[\Rightarrow D=\sqrt{18}\]
\[18\]can be written as product of \[9\]and \[2\]
\[\Rightarrow D=\sqrt{9.2}\]
Therefore \[9\] is the square of \[3\]
\[\Rightarrow D=3\sqrt{2}\]
Therefore, the distance between the two points is \[ D=3\sqrt{2}\]
Note:
Students should be very careful while doing the calculation like, \[ D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] here in this formula there itself negative signs are there students should substitute the exact values of the points with their signs, if we write \[3\] in place of \[{{x}_{1}}\] then answer will be changed we should have to write \[-3\] only.
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