The correct option is D 5 units
Since distance can never be negative, we need to find the absolute value.
(−5,0) is a point on the x-axis and it is 5 units to the left of the origin (0,0).
Alternatively, we can find the absolute value of the distance by finding the difference of the abscissas and take a modulus.
Absolute value of the distance =|−5−0|=|−5|=5 units