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b=(12))
c=(x,y)=(x,x)
liczymy boki
AB=(12-2)²+(0-0)²=100
BC=(x-12)²+(x-0)²=x²-24x+144+x²
AC=(x-2)²+(x-0)²=x²-4x+4+x²
twierdzenie pitagorasa
AB²=BC²+AC²
100=4x²-28x+148
x²-7x+12=0
Δ=1
x1=3
x2=4
x=(3,3)
y=(4,4)
to jest napewno dobrze bylo to na maturze prubnej
jak chcerz to mozesz obliczac x1,x2