Difference between revisions of "1970 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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The trisection points of <math>(-4, 5)</math> and <math>(5, -1)</math> can be found by trisecting the x-coordinates and the y-coordinates separately.  The difference of the x-coordinates is <math>9</math>, so the trisection points happen at <math>-4 + \frac{9}{3}</math> and <math>-4 + \frac{9}{3} + \frac{9}{3}</math>, which are <math>-1</math> and <math>2</math>.  Similarly, the y-coordinates have a difference of <math>6</math>, so the trisections happen at <math>3</math> and <math>1</math>.  So, the two points are <math>(-1, 3)</math> and <math>(2, 1)</math>.
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We now check which line has both <math>(3, 4)</math> and one of the two trisection points on it.  Plugging in <math>(x, y) = (3, 4)</math> into all five of the equations works.  The point <math>(2, 1)</math> doesn't work in any of the five lines.  However, (-1, 3)<math> works in line </math>E<math>.  Thus, the answer is </math>\fbox{E}$
  
 
== See also ==
 
== See also ==

Revision as of 21:26, 13 July 2019

Problem

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

$\text{(A) } 3x-2y-1=0\quad \text{(B) } 4x-5y+8=0\quad \text{(C) } 5x+2y-23=0\quad\\ \text{(D) } x+7y-31=0\quad \text{(E) } x-4y+13=0$

Solution

The trisection points of $(-4, 5)$ and $(5, -1)$ can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is $9$, so the trisection points happen at $-4 + \frac{9}{3}$ and $-4 + \frac{9}{3} + \frac{9}{3}$, which are $-1$ and $2$. Similarly, the y-coordinates have a difference of $6$, so the trisections happen at $3$ and $1$. So, the two points are $(-1, 3)$ and $(2, 1)$.

We now check which line has both $(3, 4)$ and one of the two trisection points on it. Plugging in $(x, y) = (3, 4)$ into all five of the equations works. The point $(2, 1)$ doesn't work in any of the five lines. However, (-1, 3)$works in line$E$.  Thus, the answer is$\fbox{E}$

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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