How do you use the distance formula to find the distance to the nearest tenth $J(4, - 2)$ and $U( - 2,3)?$
Answer
558.9k+ views
Hint: As we know that the distance formula is an algebraic expression that gives the distance between the pairs of the points in terms of their coordinates. It is derived by creating a triangle and using the Pythagorean theorem to find the length of the hypotenuse. The hypotenuse of the triangle is the distance between the two points. The general formula is written as $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
Complete step by step answer:
As per the given question we have the coordinates of the points $J(4, - 2)$ and $U( - 2,3).$
It gives us ${x_2} = - 2,{x_1} = 4,{y_2} = 3$ and ${y_1} = - 2$.
The distance formula is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $, by substituting the values in the formula we have $d = \sqrt {{{\left( { - 2 - 4} \right)}^2} + {{\left( {3 - ( - 2)} \right)}^2}} $.
We will now solve it further and we have $d = \sqrt {{{( - 6)}^2} + {{(3 + 2)}^2}} \Rightarrow \sqrt {36 + 25} $.
It gives us $d = \sqrt {61} $.
Hence the required value is $\sqrt {61} $ or we can say $7.8(approx)$.
Note: In the above distance formula $d$ is the distance, $\left( {{x_1},{y_1}} \right)$ are the coordinates of the first point and $\left( {{x_2},{y_2}} \right)$ are the coordinates of the second point. We should be careful while calculating the values along with the positive and negative signs. We should note that the distance between two points can never be negative and also it never decreases.
Complete step by step answer:
As per the given question we have the coordinates of the points $J(4, - 2)$ and $U( - 2,3).$
It gives us ${x_2} = - 2,{x_1} = 4,{y_2} = 3$ and ${y_1} = - 2$.
The distance formula is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $, by substituting the values in the formula we have $d = \sqrt {{{\left( { - 2 - 4} \right)}^2} + {{\left( {3 - ( - 2)} \right)}^2}} $.
We will now solve it further and we have $d = \sqrt {{{( - 6)}^2} + {{(3 + 2)}^2}} \Rightarrow \sqrt {36 + 25} $.
It gives us $d = \sqrt {61} $.
Hence the required value is $\sqrt {61} $ or we can say $7.8(approx)$.
Note: In the above distance formula $d$ is the distance, $\left( {{x_1},{y_1}} \right)$ are the coordinates of the first point and $\left( {{x_2},{y_2}} \right)$ are the coordinates of the second point. We should be careful while calculating the values along with the positive and negative signs. We should note that the distance between two points can never be negative and also it never decreases.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Which gas is abundant in air class 11 chemistry CBSE

