wyznaczyc rownanie prostej
A(1,5)
B(-3,-5)
z obliczeniami prosze. ma wyjsc ze y=5/2x+5/2 niby
A = (1; 5), B = ( -3; -5)
y = ax + b
zatem
5 = a*1 + b
-5 = a*(-3) + b
-----------------
a + b = 5
-3a + b = - 5
-------------- odejmujemy stronami
a - ( -3a) = 5 - (-5)
4a = 10 / : 4
a = 10/4 = 5/2
b = 5 - a = 5 - 5/2 = 5/2
zatem mamy
y = (5/2) x + 5/2 lub y = 2,5 x + 2,5
====================================
y = ax+b
pierwsza współrzędna to zawsze x, a druga y
(-3-1)(y-5) = (-5-5)(x-1)
(-4)*(y-5) = (-10)*(x-1)
-4y+20 = -10x+10
-4y = -10x+10-20
-4y = -10x-10 |:(-4)
y = 10/4x+10/4
y = 5/2x+5/2
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A = (1; 5), B = ( -3; -5)
y = ax + b
zatem
5 = a*1 + b
-5 = a*(-3) + b
-----------------
a + b = 5
-3a + b = - 5
-------------- odejmujemy stronami
a - ( -3a) = 5 - (-5)
4a = 10 / : 4
a = 10/4 = 5/2
b = 5 - a = 5 - 5/2 = 5/2
zatem mamy
y = (5/2) x + 5/2 lub y = 2,5 x + 2,5
====================================
y = ax+b
pierwsza współrzędna to zawsze x, a druga y
A(1,5)
B(-3,-5)
(-3-1)(y-5) = (-5-5)(x-1)
(-4)*(y-5) = (-10)*(x-1)
-4y+20 = -10x+10
-4y = -10x+10-20
-4y = -10x-10 |:(-4)
y = 10/4x+10/4
y = 5/2x+5/2