Odpowiedź:
A = ( - 2, - 2) B = ( 2, 0 ) C = ( 4, 6 )
I AB I² = ( 2 - (- 2))² + ( 0 - ( - 2))² = 4² + 2² = 16 + 4 = 20 = 4*5
więc
I AB I = [tex]\sqrt{4*5} = 2\sqrt{5}[/tex]
=======================
pr. AB y = a x + b
a = [tex]\frac{0 - ( - 2)}{2 - (-2)} = \frac{2}{4} = 0,5[/tex]
y = 0,5 x + b B = ( 2, 0)
0 = 0,5*2 + b = 1 + b
b = - 1
y = 0,5 x - 1 / * 2
2 y = x - 2
x - 2 y - 2 = 0 C = ( 4, 6 )
--------------------
Odległość C od pr. AB
h = I 1*4 - 2*6 - 2 I : [tex]\sqrt{1^2 + ( - 2)^2} = 10 : \sqrt{5} = 2\sqrt{5}[/tex]
h = 2 [tex]\sqrt{5}[/tex]
========
Pole Δ
P = 0,5* I AB I * h = 0,5* 2√5*2√5 = 2*5 = 10 j²
=========================================
Szczegółowe wyjaśnienie:
A x + B y + C = 0 P = ( [tex]x_0, y_0[/tex] )
Odległość punktu P od prostej obliczamy ze wzoru
d = I A*[tex]x_0 +[/tex] B*[tex]y_0[/tex] + C I : [tex]\sqrt{A^2 + B^2}[/tex]
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Odpowiedź:
A = ( - 2, - 2) B = ( 2, 0 ) C = ( 4, 6 )
I AB I² = ( 2 - (- 2))² + ( 0 - ( - 2))² = 4² + 2² = 16 + 4 = 20 = 4*5
więc
I AB I = [tex]\sqrt{4*5} = 2\sqrt{5}[/tex]
=======================
pr. AB y = a x + b
a = [tex]\frac{0 - ( - 2)}{2 - (-2)} = \frac{2}{4} = 0,5[/tex]
y = 0,5 x + b B = ( 2, 0)
więc
0 = 0,5*2 + b = 1 + b
b = - 1
y = 0,5 x - 1 / * 2
2 y = x - 2
x - 2 y - 2 = 0 C = ( 4, 6 )
--------------------
Odległość C od pr. AB
h = I 1*4 - 2*6 - 2 I : [tex]\sqrt{1^2 + ( - 2)^2} = 10 : \sqrt{5} = 2\sqrt{5}[/tex]
h = 2 [tex]\sqrt{5}[/tex]
========
Pole Δ
P = 0,5* I AB I * h = 0,5* 2√5*2√5 = 2*5 = 10 j²
=========================================
Szczegółowe wyjaśnienie:
A x + B y + C = 0 P = ( [tex]x_0, y_0[/tex] )
Odległość punktu P od prostej obliczamy ze wzoru
d = I A*[tex]x_0 +[/tex] B*[tex]y_0[/tex] + C I : [tex]\sqrt{A^2 + B^2}[/tex]
=================================