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Difference between revisions of "1984 AHSME Problems/Problem 10"
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The one perpenicular to <math> AB </math> and passing through <math> A </math> and the one perpendicular to <math> BC </math> and passing through <math> C </math>.
 
The one perpenicular to <math> AB </math> and passing through <math> A </math> and the one perpendicular to <math> BC </math> and passing through <math> C </math>.
 
The [[slope]] of the first line is the negative [[reciprocal]] of the slope of the line through <math> AB </math>, which, using the slope formula, is <math> \frac{1}{3} </math>, so the slope of the perpendicular line is <math> -3 </math>. It passes through <math> (1,2) </math>, so the equation of the line in point slope form is <math> y-2=-3(x-1) </math>, or <math> y=-3x+5 </math>. Similarly, the slope of the second line is <math> \frac{-1}{-3}=\frac{1}{3} </math>, and, since it passes through <math> (-1,-2) </math>, its equation is <math> y+2=\frac{1}{3}(x+1) </math>, or <math> y=\frac{1}{3}x-\frac{5}{3} </math>. To find the intersection, we have <math> -3x+5=y=\frac{1}{3}x-\frac{5}{3} </math>, and solving for <math> x </math> yields <math> x=2 </math>. Plugging this back into the equation yields <math> y=-1 </math>, so the remaining point in the Cartesian plane is <math> (2, -1) </math>, and in the complex plane is <math> 2-i, \boxed{\text{B}} </math>.
 
The [[slope]] of the first line is the negative [[reciprocal]] of the slope of the line through <math> AB </math>, which, using the slope formula, is <math> \frac{1}{3} </math>, so the slope of the perpendicular line is <math> -3 </math>. It passes through <math> (1,2) </math>, so the equation of the line in point slope form is <math> y-2=-3(x-1) </math>, or <math> y=-3x+5 </math>. Similarly, the slope of the second line is <math> \frac{-1}{-3}=\frac{1}{3} </math>, and, since it passes through <math> (-1,-2) </math>, its equation is <math> y+2=\frac{1}{3}(x+1) </math>, or <math> y=\frac{1}{3}x-\frac{5}{3} </math>. To find the intersection, we have <math> -3x+5=y=\frac{1}{3}x-\frac{5}{3} </math>, and solving for <math> x </math> yields <math> x=2 </math>. Plugging this back into the equation yields <math> y=-1 </math>, so the remaining point in the Cartesian plane is <math> (2, -1) </math>, and in the complex plane is <math> 2-i, \boxed{\text{B}} </math>.
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NICE! You can also just graph and use pythag.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=9|num-a=11}}
 
{{AHSME box|year=1984|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:54, 3 January 2021

Problem

Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i, -2+i$, and $-1-2i$. The fourth number is

$\mathrm{(A) \ }2+i \qquad \mathrm{(B) \ }2-i \qquad \mathrm{(C) \ } 1-2i \qquad \mathrm{(D) \ }-1+2i \qquad \mathrm{(E) \ } -2-i$

Solution

Perhaps the easiest way to attack this is to transfer this to the Cartesian plane. The points then would be $(1, 2), (-2, 1),$ and $(-1, -2)$, assuming the real axis was horizontal. Let these points be $A, B$ and $C$, respectively. The remaining point is then the intersection of the following perpendicular lines: The one perpenicular to $AB$ and passing through $A$ and the one perpendicular to $BC$ and passing through $C$. The slope of the first line is the negative reciprocal of the slope of the line through $AB$, which, using the slope formula, is $\frac{1}{3}$, so the slope of the perpendicular line is $-3$. It passes through $(1,2)$, so the equation of the line in point slope form is $y-2=-3(x-1)$, or $y=-3x+5$. Similarly, the slope of the second line is $\frac{-1}{-3}=\frac{1}{3}$, and, since it passes through $(-1,-2)$, its equation is $y+2=\frac{1}{3}(x+1)$, or $y=\frac{1}{3}x-\frac{5}{3}$. To find the intersection, we have $-3x+5=y=\frac{1}{3}x-\frac{5}{3}$, and solving for $x$ yields $x=2$. Plugging this back into the equation yields $y=-1$, so the remaining point in the Cartesian plane is $(2, -1)$, and in the complex plane is $2-i, \boxed{\text{B}}$.


NICE! You can also just graph and use pythag.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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