Math: Distance Formula
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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.
![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](https://image.slidesharecdn.com/distanceformula-171005084034/85/Math-Distance-Formula-4-320.jpg)
![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.](https://image.slidesharecdn.com/distanceformula-171005084034/85/Math-Distance-Formula-5-320.jpg)
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 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.