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Question: Find the distance between the points \(P\left( -6,7 \right)\) and \(Q\left( -1,-5 \right)\) ....

Find the distance between the points P(−6,7)P\left( -6,7 \right) and Q(−1,−5)Q\left( -1,-5 \right) .

Explanation

Solution

we need to find the distance between the points P(-6,7) and Q(-1,-5). These two points both lie in the XY-plane. For finding out the distance between the two points lying in the XY-plane, we use the formula (x2−x1)2+(y2−y1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} where (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) are the coordinates of the points between whose distance we want to find out. Using this formula, we will get our answer.

Complete step by step answer:
Here, we have been given the points P(-6,7) and Q(-1,-5).

Let the point P be (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and let the point Q be (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) .
Thus, we get the values of x1,x2,y1{{x}_{1}},{{x}_{2}},{{y}_{1}} and y2{{y}_{2}} as:
x1=−6 x2=−1 y1=7 y2=−5 \begin{aligned} & {{x}_{1}}=-6 \\\ & {{x}_{2}}=-1 \\\ & {{y}_{1}}=7 \\\ & {{y}_{2}}=-5 \\\ \end{aligned}
Now, we know that the distance between the two points (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) is given by the formula (x2−x1)2+(y2−y1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} .
Putting the values of x1,x2,y1{{x}_{1}},{{x}_{2}},{{y}_{1}} and y2{{y}_{2}} in this formula, we get the distance PQ as:
PQ=(−1−(−6))2+(−5−(7))2 ⇒PQ=(5)2+(−12)2 \begin{aligned} & PQ=\sqrt{{{\left( -1-\left( -6 \right) \right)}^{2}}+{{\left( -5-\left( 7 \right) \right)}^{2}}} \\\ & \Rightarrow PQ=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( -12 \right)}^{2}}} \\\ \end{aligned}
⇒PQ=25+144\Rightarrow PQ=\sqrt{25+144}
⇒PQ=169 ⇒PQ=13 \begin{aligned} & \Rightarrow PQ=\sqrt{169} \\\ & \Rightarrow PQ=13 \\\ \end{aligned}

Therefore, the distance between the points P(-6,7) and Q(-1,-5) is PQ=13unitsPQ=13units

Note: This formula is applicable for all the coordinates belonging to the XY-plane. But remember, this formula is only applicable for coordinates of the XY-plane. If we have been given a 3-dimensional coordinates, then we will have to use the formula (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} where (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and (x2,y2,z2)\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) are the 3-dimensional coordinates. This formula is similar to the formula we used here, only an extra (z2−z1)2{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}} is added inside the under root. Thus it can be remembered in the same way as we remember the formula used above.