<meta http-equiv="refresh" content="1; url=/nojavascript/"> Distance Formula ( Read ) | Algebra | CK-12 Foundation
Dismiss
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Distance Formula

%
Best Score
Practice Distance Formula
Practice
Best Score
%
Practice Now
The Distance Formula
 0  0  0

Triangle ABC has vertices A(-5, 7), B(-8, 6) and C(-3, 3) . The triangle is reflected about the y -axis to form triangle A^\prime B^\prime C^\prime . Assuming that \angle A= \angle A^\prime, \angle B= \angle B^\prime , and \angle C= \angle C^\prime , prove the two triangles are congruent.

Watch This

First watch this video to learn about the distance formula.

CK-12 Foundation Chapter10TheDistanceFormulaA

Then watch this video to see some examples.

CK-12 Foundation Chapter10TheDistanceFormulaB

Guidance

Two shapes are congruent if they are exactly the same shape and exactly the same size. In congruent shapes, all corresponding sides will be the same length and all corresponding angles will be the same measure. Translations, reflections, and rotations all create congruent shapes.

If you want to determine whether two segments are the same length, you could try to use a ruler. Unfortunately, it's hard to be very precise with a ruler. You could also use geometry software, but that is not always available. If the segments are on the coordinate plane and you know their endpoints, you can use the distance formula:

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

The distance formula helps justify congruence by proving that the sides of the preimage have the same length as the sides of the transformed image. The distance formula is derived using the Pythagorean Theorem, which you will learn more about in geometry.

Example A

Line segment AB is translated 5 units to the right and 6 units down to produce line A^\prime B^\prime . The diagram below shows the endpoints of lines AB and A^\prime B^\prime . Prove the two line segments are congruent.

Solution:

d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-4-3\right)^2+ \left(2-2\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(1-8\right)^2+ \left(-4- \left(-4\right)\right)^2} \\d_{AB}&= \sqrt{\left(-7\right)^2+ \left(0\right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(-7\right)^2+ \left(0\right)^2} \\d_{AB}&= \sqrt{49+0} && d_{A^\prime B^\prime}= \sqrt{49+0} \\d_{AB}&= \sqrt{49} && d_{A^\prime B^\prime}= \sqrt{49} \\d_{AB}&=7 \ cm && d_{A^\prime B^\prime}=7 \ cm

Example B

Line segment AB has been rotated about the origin 90^\circ CCW to produce A^\prime B^\prime . The diagram below shows the lines AB and A^\prime B^\prime . Prove the two line segments are congruent.

Solution:

d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-4-3\right)^2+ \left(2-2\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(-2- \left(-2\right)\right)^2+ \left(-4-3\right)^2} \\d_{AB}&= \sqrt{\left(-7\right)^2+ \left(0\right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(0\right)^2+ \left(-7\right)^2} \\d_{AB}&= \sqrt{49+0} && d_{A^\prime B^\prime}= \sqrt{0+49} \\d_{AB}&= \sqrt{49} && d_{A^\prime B^\prime}= \sqrt{49} \\d_{AB}&=7 \ cm && d_{A^\prime B^\prime}=7 \ cm

Example C

The square ABCD has been reflected about the line y = x to produce A^\prime B^\prime C^\prime D^\prime as shown in the diagram below. Prove the two are congruent.

Solution: Since the figures are squares, you can conclude that all angles are the same and equal to 90^\circ . You can also conclude that for each square, all the sides are the same length. Therefore, all you need to verify is that m \overline{AB}=m \overline{A^\prime B^\prime} .

d_{AB}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{AB}&= \sqrt{\left(-6.1- \left(-3\right)\right)^2+ \left(9.3-4.9\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(9.3- 4.9\right)^2+ \left(-6.1- \left(-3\right)\right)^2} \\d_{AB}&= \sqrt{\left(-3.1\right)^2+ \left(4.4\right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(4.4\right)^2+ \left(-3.1\right)^2} \\d_{AB}&= \sqrt{9.61+19.36} && d_{A^\prime B^\prime}= \sqrt{19.36+9.61} \\d_{AB}&= \sqrt{28.97} && d_{A^\prime B^\prime}= \sqrt{28.97} \\d_{AB}&=5.38 \ cm && d_{A^\prime B^\prime}=5.38 \ cm

Since m \overline{AB}=m \overline{A^\prime B^\prime} and both shapes are squares, all 8 sides must be the same length. Therefore, the two squares are congruent.

Concept Problem Revisited

To prove congruence, prove that m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AC}=m \overline{A^\prime C^\prime}, and m \overline{BC}=m \overline{B^\prime C^\prime} .

d_{AB}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AB}&= \sqrt{\left(-5- \left(-8 \right) \right)^2+ \left(7-6 \right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(5-8 \right)^2+ \left(7-6 \right)^2} \\d_{AB}&= \sqrt{\left(3 \right)^2+ \left(1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{ \left(-3 \right)^2+ \left(1 \right)^2} \\d_{AB}&= \sqrt{9+1} && d_{A^\prime B^\prime}= \sqrt{9+1} \\d_{AB}&= \sqrt{10} && d_{A^\prime B^\prime}= \sqrt{10} \\d_{AB}&=3.16 \ cm && d_{A^\prime B^\prime}=3.16 \ cm

d_{AC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime C^\prime}= \sqrt{ \left(x_2-x_1 \right)^2+ \left(y_2-y_1\right)^2} \\d_{AC}&= \sqrt{\left(-5- \left(-3 \right) \right)^2+ \left(7-3 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(5-3 \right)^2+ \left(7-3 \right)^2} \\d_{AC}&= \sqrt{\left(-2 \right)^2+ \left(4 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(2 \right)^2+ \left(4 \right)^2} \\d_{AC}&= \sqrt{4+16} && d_{A^\prime C^\prime}= \sqrt{4+16} \\d_{AC}&= \sqrt{20} && d_{A^\prime C^\prime}= \sqrt{20} \\d_{AC}&=4.47 \ cm && d_{A^\prime C^\prime}=4.72 \ cm

d_{BC}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{A^\prime C^\prime}= \sqrt{ \left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{BC}&= \sqrt{\left(-8- \left(-3\right)\right)^2+ \left(6-3\right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(8-3\right)^2+ \left(6-3\right)^2} \\d_{BC}&= \sqrt{\left(-5\right)^2+ \left(3\right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(5\right)^2+ \left(3\right)^2} \\d_{BC}&= \sqrt{25+9} && d_{A^\prime C^\prime}= \sqrt{25+9} \\d_{BC}&= \sqrt{34} && d_{A^\prime C^\prime}= \sqrt{34} \\d_{BC}&=5.83 \ cm && d_{A^\prime C^\prime}=5.83 \ cm

It is given that \angle A= \angle A^\prime, \angle B= \angle B^\prime, and \angle C= \angle C^\prime , and the distance formula proved that m \overline{AB}=m \overline{A^\prime B^\prime},m \overline{AC}=m \overline{A^\prime C^\prime}, and m \overline{BC}=m \overline{B^\prime C^\prime} . Therefore the two triangles are congruent.

Vocabulary

Congruent
Two shapes or segments are congruent if they are exactly the same shape and size.
Distance Formula
The distance formula , which finds the distance between two points, is d= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} .

Guided Practice

1. Line segment \overline{ST} drawn from S(-3, 4) to T(-3, 8) has undergone a reflection in the y -axis to produce Line S^\prime T^\prime drawn from S^\prime (3, 4) to T^\prime (4, 8) . Draw the preimage and image and prove the two lines are congruent.

2. The triangle below has undergone a rotation of 90^\circ CW about the origin. Given that all of the angles are equal, draw the transformed image and prove the two figures are congruent.

3. The polygon below has undergone a translation of 7 units to the left and 1 unit up. Given that all of the angles are equal, draw the transformed image and prove the two figures are congruent.

Answers:

1.

d_{ST}&= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(x_2-x_1\right)^2+ \left(y_2-y_1\right)^2} \\d_{ST}&= \sqrt{\left(-3- \left(-4\right)\right)^2+ \left(4-8\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(3-4\right)^2+ \left(4-8\right)^2} \\d_{ST}&= \sqrt{\left(1\right)^2+ \left(-4\right)^2} && d_{S^\prime T^\prime}= \sqrt{\left(-1\right)^2+ \left(-4\right)^2} \\d_{ST}&= \sqrt{1+16} && d_{S^\prime T^\prime}= \sqrt{1+16} \\d_{ST}&= \sqrt{17} && d_{S^\prime T^\prime}= \sqrt{17} \\d_{ST}&=4.12 \ cm && d_{S^\prime T^\prime}=4.12 \ cm

2.

d_{AB}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AB}&= \sqrt{\left(2-7 \right)^2+ \left(2-3 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(3-2 \right)^2+ \left(-7- \left(-2 \right)\right)^2} \\d_{AB}&= \sqrt{\left(-5\right)^2+ \left(-1 \right)^2} && d_{A^\prime B^\prime}= \sqrt{\left(1 \right)^2+ \left(-5 \right)^2} \\d_{AB}&= \sqrt{25+1} && d_{A^\prime B^\prime}= \sqrt{1+25} \\d_{AB}&= \sqrt{26} && d_{A^\prime B^\prime}= \sqrt{26} \\d_{AB}&=5.10 \ cm && d_{A^\prime B^\prime}=5.10 \ cm

d_{AC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{AC}&= \sqrt{\left(2-4 \right)^2+ \left(2-6 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(2-6 \right)^2+ \left(-2- \left(-4 \right)\right)^2} \\d_{AC}&= \sqrt{\left(-2 \right)^2+ \left(-4 \right)^2} && d_{A^\prime C^\prime}= \sqrt{\left(-4 \right)^2+ \left(2 \right)^2} \\d_{AC}&= \sqrt{4+16} && d_{A^\prime C^\prime}= \sqrt{16+4} \\d_{AC}&= \sqrt{20} && d_{A^\prime C^\prime}= \sqrt{20} \\d_{AC}&=4.47 \ cm && d_{A^\prime C^\prime}=4.72 \ cm

d_{BC}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{BC}&= \sqrt{\left(7-4 \right)^2+ \left(3-6 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(3-6 \right)^2+ \left(-7- \left(-4 \right)\right)^2} \\d_{BC}&= \sqrt{\left(3 \right)^2+ \left(-3 \right)^2} && d_{B^\prime C^\prime}= \sqrt{\left(-3 \right)^2+ \left(-3 \right)^2} \\d_{BC}&= \sqrt{9+9} && d_{B^\prime C^\prime}= \sqrt{9+9} \\d_{BC}&= \sqrt{18} && d_{B^\prime C^\prime}= \sqrt{18} \\d_{BC}&=4.24 \ cm && d_{B^\prime C^\prime}=4.24 \ cm

3.

d_{DE}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{DE}&= \sqrt{\left(3.5-6 \right)^2+ \left(1-3 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(-3.5- \left(-1 \right) \right)^2+ \left(2-4 \right)^2} \\d_{DE}&= \sqrt{\left(-2.5 \right)^2+ \left(-2 \right)^2} && d_{D^\prime E^\prime}= \sqrt{\left(-2.5 \right)^2+ \left(-2 \right)^2} \\d_{DE}&= \sqrt{6.25+4} && d_{D^\prime E^\prime}= \sqrt{6.25+4} \\d_{DE}&= \sqrt{10.25} && d_{D^\prime E^\prime}= \sqrt{10.25} \\d_{DE}&=3.20 \ cm && d_{D^\prime E^\prime}=3.20 \ cm

d_{EF}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{EF}&= \sqrt{\left(6-5 \right)^2+ \left(3-6 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(-1- \left(-2 \right) \right)^2+ \left(4-7 \right)^2} \\d_{EF}&= \sqrt{\left(1 \right)^2+ \left(-3 \right)^2} && d_{E^\prime F^\prime}= \sqrt{\left(1 \right)^2+ \left(-3 \right)^2} \\d_{EF}&= \sqrt{1+9} && d_{E^\prime F^\prime}= \sqrt{1+9} \\d_{EF}&= \sqrt{10} && d_{E^\prime F^\prime}= \sqrt{10} \\d_{EF}&=3.16 \ cm && d_{E^\prime F^\prime}=3.16 \ cm

d_{FG}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{FG}&= \sqrt{\left(5-2 \right)^2+ \left(6-6 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(-2- \left(-5 \right) \right)^2+ \left(7-7 \right)^2} \\d_{FG}&= \sqrt{\left(3 \right)^2+ \left(0 \right)^2} && d_{F^\prime G^\prime}= \sqrt{\left(3 \right)^2+ \left(0 \right)^2} \\d_{FG}&= \sqrt{9+0} && d_{F^\prime G^\prime}= \sqrt{9+0} \\d_{FG}&= \sqrt{9} && d_{F^\prime G^\prime}= \sqrt{9} \\d_{FG}&=3.00 \ cm && d_{F^\prime G^\prime}=3.00 \ cm

d_{GH}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{GH}&= \sqrt{\left(2-1 \right)^2+ \left(6-3 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(-5- \left(-6 \right) \right)^2+ \left(7-4 \right)^2} \\d_{GH}&= \sqrt{\left(1 \right)^2+ \left(3 \right)^2} && d_{G^\prime H^\prime}= \sqrt{\left(1 \right)^2+ \left(3 \right)^2} \\d_{GH}&= \sqrt{1+9} && d_{G^\prime H^\prime}= \sqrt{1+9} \\d_{GH}&= \sqrt{10} && d_{G^\prime H^\prime}= \sqrt{10} \\d_{GH}&=3.16 \ cm && d_{G^\prime H^\prime}=3.16 \ cm

d_{HD}&= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(x_2-x_1 \right)^2+ \left(y_2-y_1 \right)^2} \\d_{HD}&= \sqrt{\left(1-3.5 \right)^2+ \left(3-1 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(-6- \left(-3.5 \right) \right)^2+ \left(4-2 \right)^2} \\d_{HD}&= \sqrt{\left(-2.5 \right)^2+ \left(2 \right)^2} && d_{H^\prime D^\prime}= \sqrt{\left(-2.5 \right)^2+ \left(2 \right)^2} \\d_{HD}&= \sqrt{6.25+4} && d_{H^\prime D^\prime}= \sqrt{6.25+4} \\d_{HD}&= \sqrt{10.25} && d_{H^\prime D^\prime}= \sqrt{10.25} \\d_{HD}&=3.20 \ cm && d_{H^\prime D^\prime}=3.20 \ cm

Practice

Find the length of each line segment below given its endpoints. Leave all answers in simplest radical form.

  1. Line segment AB given A(5, 7) and B(3, 9) .
  2. Line segment BC given B(3, 8) and C(5, 2) .
  3. Line segment CD given C(4, 6) and D(3, 5) .
  4. Line segment DE given D(9, 11) and E(2, 2) .
  5. Line segment EF given E(1, 1) and F(8, 7) .
  6. Line segment FG given F(3, 6) and G(2, 4) .
  7. Line segment GH given G(-2, 4) and H(6, -1) .
  8. Line segment HI given H(1, -5) and I(3, 3) .
  9. Line segment IJ given I(3.4, 7) and J(1, 6) .
  10. Line segment JK given J(6, -3) and K(-2, 4) .
  11. Line segment KL given K(-3, -3) and L(2, -1) .

For each of the diagrams below, assume the corresponding angles are congruent. Find the lengths of the line segments to prove congruence.

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...

Original text


ShareThis Copy and Paste