How do you use the distance formula to find the distance to the nearest tenth $J(4, - 2)$ and $U( - 2,3)?$
Answer
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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.
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