What is the least possible distance between a point on the line y = (3/4)x − 3 and a point on the circle centered at the origin with a radius of 1?
1.4
sqrt(2)
1.7
sqrt(3)
2.0
I've changed the wording of the problem so that the solution does not require knowledge of the equation of a circle.
DRAW the figure:
The SHORTEST DISTANCE between a point and a line must form a RIGHT ANGLE with the line.
Thus:
AC = the shortest distance between the center of the circle and y = (3/4)x - 3.
EC = the shortest distance between a point on the circle and y = (3/4)x - 3.
∆ABD is a 3-4-5 triangle.
Area of ∆ABD = (1/2)(AD)(AB) = (1/2)(3)(4) = 6.
If we consider BD the base of ∆ABD, then AC is the corresponding height.
Since 1/2 the product of any base and height must yield the same area, we get:
(1/2)(BD)(AC) = 6
(1/2)(5)(AC) = 6.
AC = 12/5.
Since AE is radius of the circle, AE = 1.
Thus, EC = AC - AE = 12/5 - 1 = 7/5 = 1.4.
The correct answer is
A.
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