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1 of 6
THE
DISTANCE
FORMULA
A
B
C
(-6, -3)
(6, -3)
(6, 2)
Example:
Coordinates of A and B are (-6, -3)
and (6, 2), respectively. How long
is 𝐴𝐡 ?
Solution:
A vertical line through B and a
horizontal line through A are drawn.
These lines intersect at C (6, -3).
AC = |-6 -6| =|-12| = 12
BC = |2- -3| = |5|= 5
𝐴𝐡2=𝐴𝐢2 + 𝐡𝐢2
AB = 𝐴𝐢2 + 𝐡𝐢2
AB = 122 + 52
= 144 + 2
= 169
AB = 13
1313
A
B
C
(𝒙 𝟏, π’š 𝟏) (𝒙 𝟐, π’š 𝟏)
(𝒙 𝟐, π’š 𝟐)
The coordinates of C are (π‘₯2, 𝑦1)
and
AC =|π‘₯2 βˆ’ π‘₯1|, BC = |𝑦2 βˆ’ 𝑦1|
Any horizontal line is βŠ₯ to any
vertical line.
Thus, Δ𝐴𝐡𝐢 is a right triangle.
𝐴𝐡2
= 𝐴𝐢2
+𝐡𝐢2
AB = 𝐴𝐢2 + 𝐡𝐢2
AB = |π‘₯2 βˆ’ π‘₯1|2 + |𝑦2 βˆ’ 𝑦1|2
Since |π‘₯2 βˆ’ π‘₯1|2 = (π‘₯2 + π‘₯1)2, we have
AB = (π‘₯2βˆ’π‘₯1)2 + (𝑦2+𝑦1)2
𝒙 𝟏
𝒙 𝟐
π’š 𝟐
π’š 𝟏
y
x
|𝒙 𝟐 βˆ’ 𝒙 𝟏|
|π’š 𝟐 βˆ’ π’š 𝟏|
Example 2
Determine XY if X(-4, 5) and
Y(1,2).
XY= (π‘₯2βˆ’π‘₯1)2 + (𝑦2+𝑦1)2
XY= [1 βˆ’ βˆ’4 ]2+(2 βˆ’ 5)2
XY = (5)2+(βˆ’3)2
XY = 25 + 9
XY= 34
10
8
6
4
2
-4 -2 2 4 6
X
Y
M
C
A
Example 3
Prove that the triangle whose vertices
are A (-1, -1), C(3, -4) and M (2, 3) is
isosceles.
Proof:
AC = (βˆ’1 βˆ’ 3)2+[βˆ’1 βˆ’ βˆ’4 ]2
= (βˆ’4)2+(3)2
= 16 + 9
AC = 5
AM = (βˆ’1 βˆ’ 2)2+(βˆ’1 βˆ’ 3)2
= (βˆ’3)2+(βˆ’4)2
= 9 + 16
AM = 5
CM= (3 βˆ’ 2)2+(βˆ’4 βˆ’ 3)2
= (1)2+(βˆ’7)2
= 1 + 49
= 50
CM = 5 2
Since AC = AM, then 𝐴𝐢 β‰… 𝐴𝑀 and Δ𝐴𝐢𝑀
is an isosceles triangle.
SGZMJ XNT!THANK YOU!

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Introduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher EducationIntroduction to ArtificiaI Intelligence in Higher Education
Introduction to ArtificiaI Intelligence in Higher Education
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Math: Distance Formula

  • 2. A B C (-6, -3) (6, -3) (6, 2) Example: Coordinates of A and B are (-6, -3) and (6, 2), respectively. How long is 𝐴𝐡 ? Solution: A vertical line through B and a horizontal line through A are drawn. These lines intersect at C (6, -3). AC = |-6 -6| =|-12| = 12 BC = |2- -3| = |5|= 5 𝐴𝐡2=𝐴𝐢2 + 𝐡𝐢2 AB = 𝐴𝐢2 + 𝐡𝐢2 AB = 122 + 52 = 144 + 2 = 169 AB = 13 1313
  • 3. A B C (𝒙 𝟏, π’š 𝟏) (𝒙 𝟐, π’š 𝟏) (𝒙 𝟐, π’š 𝟐) The coordinates of C are (π‘₯2, 𝑦1) and AC =|π‘₯2 βˆ’ π‘₯1|, BC = |𝑦2 βˆ’ 𝑦1| Any horizontal line is βŠ₯ to any vertical line. Thus, Δ𝐴𝐡𝐢 is a right triangle. 𝐴𝐡2 = 𝐴𝐢2 +𝐡𝐢2 AB = 𝐴𝐢2 + 𝐡𝐢2 AB = |π‘₯2 βˆ’ π‘₯1|2 + |𝑦2 βˆ’ 𝑦1|2 Since |π‘₯2 βˆ’ π‘₯1|2 = (π‘₯2 + π‘₯1)2, we have AB = (π‘₯2βˆ’π‘₯1)2 + (𝑦2+𝑦1)2 𝒙 𝟏 𝒙 𝟐 π’š 𝟐 π’š 𝟏 y x |𝒙 𝟐 βˆ’ 𝒙 𝟏| |π’š 𝟐 βˆ’ π’š 𝟏|
  • 4. Example 2 Determine XY if X(-4, 5) and Y(1,2). XY= (π‘₯2βˆ’π‘₯1)2 + (𝑦2+𝑦1)2 XY= [1 βˆ’ βˆ’4 ]2+(2 βˆ’ 5)2 XY = (5)2+(βˆ’3)2 XY = 25 + 9 XY= 34 10 8 6 4 2 -4 -2 2 4 6 X Y
  • 5. M C A Example 3 Prove that the triangle whose vertices are A (-1, -1), C(3, -4) and M (2, 3) is isosceles. Proof: AC = (βˆ’1 βˆ’ 3)2+[βˆ’1 βˆ’ βˆ’4 ]2 = (βˆ’4)2+(3)2 = 16 + 9 AC = 5 AM = (βˆ’1 βˆ’ 2)2+(βˆ’1 βˆ’ 3)2 = (βˆ’3)2+(βˆ’4)2 = 9 + 16 AM = 5 CM= (3 βˆ’ 2)2+(βˆ’4 βˆ’ 3)2 = (1)2+(βˆ’7)2 = 1 + 49 = 50 CM = 5 2 Since AC = AM, then 𝐴𝐢 β‰… 𝐴𝑀 and Δ𝐴𝐢𝑀 is an isosceles triangle.