1961 AMC 12 Problem 3

Attempt Problem 3 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

If the graphs of

2y+x+3=0

and

3y+ax+2=0

are to meet at right angles, the value of

a

is:

\pm\dfrac23

-\dfrac23

-\dfrac32

6

-6

Put both equations into slope-intercept form

The product of the slopes of perpendicular nonvertical lines is

-1

The two slopes are

-\frac{1}{2}

and

-\frac{a}{3}.

Perpendicularity requires

\left(-\frac12\right)\left(-\frac a3\right)=-1,

a=-6.

Therefore, the correct answer is E.

Problem 3 in other years