1. The document contains worked solutions to geometry problems involving distances between points and lines. 2. It is determined that the figure formed by the points is a square based on all distances being equal. 3. A second problem finds the coordinates of a point C that satisfies two linear equations, and then calculates the distance between C and another point. 4. A third problem finds the x-intercept of a line between two points, and a fourth problem finds the slope and angle between two lines using their equations.

1. CONTROL COLABORATIVO
N° 1
PFM

2. PREGUNTA 1
d(AB) = d(BC)
(4-1)^2 + (6-2)^2 = (8-4)^2 + (3-6)^2
3^2 + 4^2 = 4^2 + (-3)^2
25 = 25
d(BC) = d(CD)
(8-4)^2 + (3-6)^2 = (5-3)^2 + (-1-3)^2
4^2 + (-3)^2 = (-3)^2 + (-4)^2
25 = 25
C
(8;3)
B
(4;6)
D (5;-
1)
A
(1;2)
d
d
d
d

3. PREGUNTA 1
d(CD) = d(DA)
(5-8)^2 + (-1-3)^2 = (5-1)^2 + (-1-2)^2
(-3)^2 + (-4)^2 = 4^2 + (-3)^2
25 = 25
d(DA) = d(AB)
(5-1)^2 + (-1-2)^2 = (4-1)^2 + (6-2)^2
4^2 + (-3)^2 = (3)^2 + (4)^2
25 = 25
C
(8;3)
B
(4;6)
D (5;-
1)
A
(1;2)
d
d
d
d

4. PREGUNTA 1
Como todos las distancias son iguales, se
puede inferir que la figura es un cuadrado.
C
(8;3)
B
(4;6)
D (5;-
1)
A
(1;2)
d
d
d
d

5. PREGUNTA 2
A) d(A;(x,y)) = d ((x;y); Chiclayo)
(x+1;74)^2 + (y-3;43)^2 = (x-0)^2 + (y-0)^2
x^2 + 3;48x + 3,0276 + y^2 – 6,86y + 11, 7649 = x^2 + y^2
3, 48x – 6,86y + 14, 7925 = 0

6. PREGUNTA 2
B) C (x ; y)
L1: 1,15x + 0,06y + 1,8 = 0
L2: -0,94x – 1,95y + 2,85 = 0
1,15 (2,86 – 1,95y/0,94) + 0,06y +
1,8 = 0
4,981 = 2,8161y
2,28 = y
X = 1,95y – 2,86 / -0,94
X = -1,68
C (-1,68 ; 2,28)

7. PREGUNTA 2
C)
(-1,68 – 0,26)^2 + (2,28 – 1,34)^2
3, 7636 + 0, 8836
2,16 x 50
107, 8 KM

8. PREGUNTA 3
A (-2 ; 1) ^ B (3 ; 1)
B (-
2,1) B
(3,1)
X = -2
m (AB) =
0
L1: x = -
2
D (AB) = (2x-3)^2 + (1+x)^2
5^2 = 4x^2 - 12x + 9 + 1 + 2x +
x^2
25 = 5x^2 - 10x + 10
0 = 5x^2 – 10x – 15
5x 5
x -3
5x + 5 = 0 x – 3
= 0
5x = -5 x = 3
X = -1

9. PREGUNTA 4
L1: x/(-2/3) + y /2 = 1
L1: -3x/x + y/2 = 1
L1: -3x+y/2 = 1
L1: -3x + y = 2
L1: 3x – y = -2
L
2
L
1ml1(
3)ml2(-
2)

10. PREGUNTA 4 Tabuland
o
L
2
L
1ml1(
3)ml2(-
2)
-2/3
x y
0 2
-2/3 0
2
m de L1
-A/B = -3/-1 = 3
mL1 = 3

11. PREGUNTA 4
L
2
L
1ml1(
3)ml2(-
2)
-2/3
2
Tan de Ω = m – m0 / 1 + m*m0
Tan de Ω = -2 – 3 / 1 + (3)(-2)
Tan de Ω = -5/-5 = 1
Tan de Ω = tan de 45°
Ω = 45°
Ω