1970 AMC 12 Problem 15

Attempt Problem 15 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

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15.

Lines in the

xy

-plane are drawn through the point

(3,4)

and the trisection points of the line segment joining the points

(-4,5)

and

(5,-1).

One of these lines has the equation:

3x-2y-1=0

4x-5y+8=0

5x+2y-23=0

x+7y-31=0

x-4y+13=0

Find the two trisection points by moving one-third and two-thirds of the displacement vector

The trisection points are

(-1,3)

and

(2,1)

; find the line from either one to

(3,4)

The displacement from

(-4,5)

(5,-1)

(9,-6).

The trisection points are therefore

(-4,5)+\tfrac13(9,-6)=(-1,3)

and

(2,1).

The line through

(-1,3)

and

(3,4)

has slope

\frac{1}{4},

y-4=\frac14(x-3),

x-4y+13=0.

Therefore, the correct answer is E.

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