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 π΄π΅ ?
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
π΄π΅2=π΄πΆ2 + π΅πΆ2
AB = π΄πΆ2 + π΅πΆ2
AB = 122 + 52
= 144 + 2
= 169
AB = 13
1313
3. A
B
C
(π π, π π) (π π, π π)
(π π, π π)
The coordinates of C are (π₯2, π¦1)
and
AC =|π₯2 β π₯1|, BC = |π¦2 β π¦1|
Any horizontal line is β₯ to any
vertical line.
Thus, Ξπ΄π΅πΆ is a right triangle.
π΄π΅2
= π΄πΆ2
+π΅πΆ2
AB = π΄πΆ2 + π΅πΆ2
AB = |π₯2 β π₯1|2 + |π¦2 β π¦1|2
Since |π₯2 β π₯1|2 = (π₯2 + π₯1)2, we have
AB = (π₯2βπ₯1)2 + (π¦2+π¦1)2
π π
π π
π π
π π
y
x
|π π β π π|
|π π β π π|
4. Example 2
Determine XY if X(-4, 5) and
Y(1,2).
XY= (π₯2βπ₯1)2 + (π¦2+π¦1)2
XY= [1 β β4 ]2+(2 β 5)2
XY = (5)2+(β3)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 = (β1 β 3)2+[β1 β β4 ]2
= (β4)2+(3)2
= 16 + 9
AC = 5
AM = (β1 β 2)2+(β1 β 3)2
= (β3)2+(β4)2
= 9 + 16
AM = 5
CM= (3 β 2)2+(β4 β 3)2
= (1)2+(β7)2
= 1 + 49
= 50
CM = 5 2
Since AC = AM, then π΄πΆ β π΄π and Ξπ΄πΆπ
is an isosceles triangle.