The document explains the distance formula for calculating the distance between points on a coordinate plane. It provides examples of how to compute distances using vertices of triangles and demonstrates the properties of isosceles triangles. The document includes specific coordinate examples and the step-by-step methodology for solving them.

1. THE
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 𝐴𝐵 ?
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
𝐴𝐵2=𝐴𝐶2 + 𝐵𝐶2
AB = 𝐴𝐶2 + 𝐵𝐶2
AB = 122 + 52
= 144 + 2
= 169
AB = 13
1313

3. A
B
C
(𝒙 𝟏, 𝒚 𝟏) (𝒙 𝟐, 𝒚 𝟏)
(𝒙 𝟐, 𝒚 𝟐)
The coordinates of C are (𝑥2, 𝑦1)
and
AC =|𝑥2 − 𝑥1|, BC = |𝑦2 − 𝑦1|
Any horizontal line is ⊥ to any
vertical line.
Thus, Δ𝐴𝐵𝐶 is a right triangle.
𝐴𝐵2
= 𝐴𝐶2
+𝐵𝐶2
AB = 𝐴𝐶2 + 𝐵𝐶2
AB = |𝑥2 − 𝑥1|2 + |𝑦2 − 𝑦1|2
Since |𝑥2 − 𝑥1|2 = (𝑥2 + 𝑥1)2, we have
AB = (𝑥2−𝑥1)2 + (𝑦2+𝑦1)2
𝒙 𝟏
𝒙 𝟐
𝒚 𝟐
𝒚 𝟏
y
x
|𝒙 𝟐 − 𝒙 𝟏|
|𝒚 𝟐 − 𝒚 𝟏|

4. Example 2
Determine XY if X(-4, 5) and
Y(1,2).
XY= (𝑥2−𝑥1)2 + (𝑦2+𝑦1)2
XY= [1 − −4 ]2+(2 − 5)2
XY = (5)2+(−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 = (−1 − 3)2+[−1 − −4 ]2
= (−4)2+(3)2
= 16 + 9
AC = 5
AM = (−1 − 2)2+(−1 − 3)2
= (−3)2+(−4)2
= 9 + 16
AM = 5
CM= (3 − 2)2+(−4 − 3)2
= (1)2+(−7)2
= 1 + 49
= 50
CM = 5 2
Since AC = AM, then 𝐴𝐶 ≅ 𝐴𝑀 and Δ𝐴𝐶𝑀
is an isosceles triangle.

6. SGZMJ XNT!THANK YOU!