Answer:
B
Explanation:
y=|x-3|
y=1/2 x
|x-3|=1/2x
if |x|=a
x=±a
x-3=±1/2 x
when x-3=1/2x
2x-6=x
x=6
so y=6/2=3
point is (6,3)
when x-3=-1/2 x
2x-6=-x
3x=6
x=6/3=2
y=1/2 x=1/2×2=1
point is (2,1)
distance=√[(2-6)²+(1-3)²]=√[16+4]=√20
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