Question

1) Oblicz pole oraz obwód równoległoboku ABCD wiedząc że: A=(-3,4), B=(2,5), C=(4,9).


2) Oblicz odleglosc punktu (-5,12) od prostej przechodzacej przez punkty (3,5) oraz (-1,4).


3) Oblicz pole kwadratu ABCD wiedzac, ze A=(4,-1) oraz C=(-3,8).

Answer 1

1.P- 4j x 5j = 20j(kwadratowych)
obw- 4j + 4j+5j+5j=18j(kwadratowych)

2. 9j

3.B=(4,8) D=(-3,-1)

P- 7jx9j=63j(kwadratowych)

Answer 2

Z.1
A =(-3;4), B = (2; 5), C = (4; 9)
Wektory AB oraz DC muszą być równe.

D = (x; y)
-->
AB = [2-(3); 5 - 4] = [ 5; 1]
-->
DC = [4 -x; 9 - y]
zatem 4 - x = 5 oraz 9 - y = 1
czyli x = 4 - 5 = - 1 oraz y = 9 - 1 = 8
D = (-1; 8 )
I AB I² = 5² + 1² = 25 + 1 = 26
I AB I = √26 oraz I CD I = √26
-->
BC = [4 - 2; 9 -5] = [ 2; 4]
I BC I² = 2² + 4² = 4 + 16 = 20
I BC I = √20 = 2√5 oraz I AD I = 2√5
L - obwód ABCD
L = 2* I AB I + 2* I BC I = 2*√26 + 2*2√5 = 2√26 + 4√5
================================================
pr AB
y = ax + b
4 = -3a + b
5 = 2a + b
------------------ odejmujemy stronami
5 - 4 = 2a -(-3a)
1 = 5a ---> a = 1/5
b = 5 - 2a = 5 - 2*(1/5) = 5 - 2/5 = 25/5 - 2/5 = 23/5
y = (1/5) x + 23/5
-----------------------
pr CD równoległa do pr AB
y =(1/5)x + b1 oraz C = (4; 9)
9 = (1/5)*4 = b1 --> b1 = 9 - 4/5 = 45/5 - 4/5 = 41/5
y = (1/5) x + 41/5
-------------------------
pr BE prostopadla do pr AB
a*a1 = - 1
(1/5)*a1 = - 1 ---> a1 = - 5
y = -5 x + b2 oraz B = (2; 5)
5 = -5*2 + b2 ----> b2 = 5 + 10 = 15
y = - 5 x + 15
-------------------- Szukam punktu E
pr CD i pr BE
y = (1/5) x + 41/5 oraz y = - 5 x + 15
(1/5)x + 41/5 = - 5x + 15 / * 5
x + 41 = -25 x + 75
x + 25 x = 75 - 41
26x = 34 --> x = 34/26 = 17/13
x = 17/13
y = -5*(17/13) + 15 = -85/13 + 15 = -85/13 + 195/13 = 110/13
y = 110/13
E = (17/13; 110/13)
h = I BE I
I BE I² = (17/13 - 2)² + (110/13 - 5)² =
= (17/13 - 26/13)² + (110/13 - 65/13)² =
= (- 9/13)² + (45/13² = 81/169 + 2025/169 = 2106/169 = 162/13
h = I BE I = √162/ √13
P = I AB I * h = √26*[√162/ √13] = [√2*√13] *[√162/ √13] =
= √2*√162 = √324 = 18
P = 18 j²
Długi sposób - bez korzystania z wzoru na odległość punktu
od prostej.
============================
z.2
P = (-5; 12)
A = ( 3; 5) , B = (-1 ; 4)
pr AB
y = ax + b
5 = 3a + b
4 = -1 a + b
------------------ odejmujemy stronami
5 - 4 = 3a -(-a) = 4a
4a = 1 --> a = 1/4
b = 4 + a = 4 + 1/4 = 17/4
y = (1/4) x + 17/4
-------------------------
pr PC prostopadła do pr AB
(1/4)*a1 = -1
a1 = -4 oraz P = (-5; 12)
y = -4 x + b
12 = -4*(-5) + b
b = 12 -20 = - 8
y = -4 x - 8
-------------------
Szukam punktu C
(1/4) x + 17/4 = - 4 x - 8 / *4
x + 17 = - 16 x - 32
x + 16 x = - 32 - 17
17 x = - 49
x = - 49/17
y = -4*(-49/17) - 8 = 196/17 - 136/17 = 60/17
C = ( -49/17; 60/17)
d = I PC I
I PC I² = (-49/17 +5)² + (60/17 - 12)² = (-49/17 + 85/17)² +
+ (60/17 - 204/17)² = (36/17)² + (144/17)² =
= 1296/289 + 20 736/289 = 22 032/289 = 1296/17
I PC I = √1296 : √17 = 36 /√17 = (36 √17) / 17
Odp. d = (36√17)/ 17
==========================
d ≈ 8,73
===========================================================
z.3
A = (4; -1) , C = (-3; 8)
I AC I² = ( -3 - 4)² + (8 - (-1))² = (-7)² + 9² = 49 + 81 = 130
I AC I = √130
P = (1/2) I AC I² = (1/2) *(√130)² = (1/2) *130 = 65 j²
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