The Distance Formula

The Distance Formula exercise

• Explain how to calculate the distance between two points.

  Hints

  The Pythagorean theorem states that the sum of the squares of the legs of any right triangle is the same as the square of the hypotenuse.

  In the picture above, the lengths of both legs are given by the difference of the coordinates of the two points.

  We don't want to know the squared distance.

  Solution

  To determine the distance between two given points, $(x_1,y_1)$ and $(x_2,y_2)$, in a coordinate plane we first draw a right angle.

  Then we use the Pythagorean theorem to get

  $(x_2-x_1)^2+(y_2-y_1)^2=c^2$,

  where $c$ is the length of the hypotenuse.

  Because the length of the hypotenuse is the desired distance, we replace it by $d$:

  $(x_2-x_1)^2+(y_2-y_1)^2=d^2$.

  Lastly, we take the square root on both sides and change the sides to get the distance formula

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

• Calculate the distance from Carlos' village to his school.

  Hints

  Here you see the distance formula for two points in a coordinate plane.

  You get the points by just looking at the coordinate plane on the map pictured above.

  For Carlos' village:

  • Draw a line parallel to the $x$-axis passing this point. The $y$-coordinate is the intersection of this line and the $y$-axis.
  • Draw a line parallel to the $y$-axis passing this point. The $x$-coordinate is the intersection of this line and the $x$-axis.

  The coordinate for Carlos' school can be found in a similar fashion.

  PEMDAS means the order of operations:

  • Paranthesis
  • Exponent
  • Multiplication
  • Division
  • Addition
  • Subtraction

  Solution

  The coordinates of Carlos' village as well as his school are:

  • Home: $(100,100)$
  • School: $(200,400)$

  The distance formula for two points in a coordinate plane is given by

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$,

  where, in our situation, we have that:

  • $x_2=200$
  • $x_1=100$
  • $y_2=400$
  • $y_1=100$

  Putting those values into the distance formula above, we get:

  $d=\sqrt{(200-100)^2+(400-100)^2}$.

  Using PEMDAS, we get

  $d=\sqrt{(100)^2+(300)^2}=\sqrt{10000+90000}=\sqrt{100000}$.

  We can simplify this expression to get

  $d=100\sqrt{10}\approx316.23$.

• Connect the ordered pairs with the right formula.

  Hints

  Pay attention to the correct order of the points.

  Just check the exponents.

  Here you see an example of using the distance formula.

  So we have $d=\sqrt{(56-12)^2+(78-34)^2}$.

  Solution

  The distance between two points is the square root of the sum of the squares of the differences of the coordinates:

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

  Let's practice it with a few pairs of points:

  The distance between $(10,20)$ and $(30,40)$ is

  $d=\sqrt{(30-10)^2+(40-20)^2}$.

  Pay attention to the order: $(10,20)$ and $(40,30)$ leads to

  $d=\sqrt{(40-10)^2+(30-20)^2}$.

  It looks quite similar, but it's totally different.

  The distance between $(40,20)$ and $(10,50)$ is

  $d=\sqrt{(10-40)^2+(50-20)^2}$,

  while the distance between $(40,20)$ and $(50,10)$ is

  $d=\sqrt{(50-40)^2+(10-20)^2}$.

• Determine the distance.

  Hints

  Use the distance formula for two points in a coordinate plane:

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

  If you have to round to the nearest hundredth, you just have to look at the third position after the decimal: if this position is less or equal $4$ then round down, otherwise round up.

  Here are two examples for rounding decimals:

  Solution

  The point representing Carlos' location is $(100,100)$ and the point representing where his friend lives is $(150,300)$.

  We use the distance formula for two points in a coordinate plane:

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$,

  with the given coordinates above. So we put them into this formula to get

  $d=\sqrt{(150-100)^2+(300-100)^2}$.

  Using PEMDAS, we get

  $d=\sqrt{(50)^2+(200)^2}=\sqrt{2500+40000}=\sqrt{42500}$.

  We can write this result as

  $d=\sqrt{2500\times17}=50\sqrt{17}$,

  or as a rounded decimal

  $d\approx206.16$.

• Find the right distance formula.

  Hints

  Here you see a right triangle with the points $(x_1,y_1)$ and $(x_2,y_2)$.

  Use the Pythagorean property:

  The sum of the squares of the legs of any right triangle is the same as the square of the hypotenuse.

  The distance is the length of the hypotenuse.

  Solution

  Two points, $(x_1,y_1)$ as well as $(x_2,y_2)$, of a right triangle are given.

  With the Pythagorean theorem, we get

  $(x_2-x_1)^2+(y_2-y_1)^2=c^2$,

  where $c$ is the length of the hypotenuse and $|x_2-x_1|$ and $|y_2-y_1|$ the lengths of the legs.

  Because the length of the hypotenuse is the desired distance, we replace it by $d$:

  $(x_2-x_1)^2+(y_2-y_1)^2=d^2$.

  Lastly, we take the square root on both sides and change the sides to get the distance formula

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

• Examine the distances of the given points.

  Hints

  Use the distance formula:

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

  Here you see an example for the calculation of the distance of the two points $(10,40)$ and $(40,20)$.

  Be careful with the order of the coordinates.

  Solution

  The distance between two points is given as the square root of the sum of the squares of the differences of the coordinates:

  $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

  Let's practice the calculation of the distance for a few pairs of points:

  $(10,20)$ and $(30,40)$ gives

  $\begin{array}{rcl} d&=&~\sqrt{(30-10)^2+(40-20)^2}\\ &=&~\sqrt{(20)^2+(20)^2}\\ &=&~\sqrt{800}=20\sqrt2 \end{array}$

  Pay attention to the order of the coordinates: $(10,20)$ and $(40,30)$ leads to

  $\begin{array}{rcl} d&=&~\sqrt{(40-10)^2+(30-20)^2}\\ &=&~\sqrt{(30)^2+(10)^2}\\ &=&~\sqrt{1000}=10\sqrt{10} \end{array}$

  So you see, it's very important to be careful with the order of the coordinates. Those calculations look quite similar, but the result is totally different.

  The distance between $(10,30)$ and $(30,30)$ is

  $\begin{array}{rcl} d&=&~\sqrt{(30-10)^2+(30-30)^2}\\ &=&~\sqrt{(20)^2}\\ &=&~20 \end{array}$

  Lastly, the distance between $(40,40)$ and $(10,90)$ is

  $\begin{array}{rcl} d&=&~\sqrt{(10-40)^2+(90-40)^2}\\ &=&~\sqrt{(-30)^2+(50)^2}\\ &=&~\sqrt{2500}=50 \end{array}$