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Difference between revisions of "2014 AMC 10A Problems/Problem 14"

(→‎See Also: problem no. incorrect)
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==Solution==
 
==Solution==
 +
<asy>//Needs refining
 +
size(20cm);
 +
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
 +
for(int i=-2;i<=8;i+=1)
 +
  draw((i,-12)--(i,12),grey);
 +
for(int j=-12;j<=12;j+=1)
 +
  draw((-2,j)--(8,j),grey);
 +
draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
 +
draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis
 +
dot((0,0));
 +
dot((6,8));
 +
draw((-2,10.66667)--(8,7.33333),Arrows);
 +
draw((7.33333,12)--(-0.66667,-12),Arrows);
 +
draw((6,8)--(0,8));
 +
draw((6,8)--(0,0));
 +
draw(rightanglemark((0,10),(6,8),(0,-10),20));
 +
label("$A$",(6,8),NE);
 +
</asy>
 +
Note that the <math>y</math>-coordinates of <math>P</math> and <math>Q</math> must be negations of each other; hence it follows that <math>O=(0,0)</math> is the midpoint of <math>\overline{PQ}</math>. Because <math>\triangle APQ</math> is right, the median <math>\overline{OA}</math> is its circumradius and <math>O</math> its circumcentre. The [[Pythagorean Theorem]] gives us <math>AO=\sqrt{6^2+8^2}=10</math>, so <math>OP=OQ=10</math>. Thus <math>P</math> has <math>y</math>-coordinate 10 and <math>Q</math> <math>y</math>-coordinate -10 and <math>PQ=20</math>. The altitude from <math>A</math> to <math>\overline{PQ}</math> is <math>6</math> (<math>\overline{PQ}</math> is on the <math>y</math>-axis, hence this altitude is just the <math>x</math>-coordinate of <math>A</math>), so <math>[\triangle APQ]=\dfrac{20\cdot6}{2}=60\implies\boxed{\textbf{(D)}\ 60}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 07:33, 7 February 2014

Problem

The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?

$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution

[asy]//Needs refining size(20cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1) draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1) draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); [/asy] Note that the $y$-coordinates of $P$ and $Q$ must be negations of each other; hence it follows that $O=(0,0)$ is the midpoint of $\overline{PQ}$. Because $\triangle APQ$ is right, the median $\overline{OA}$ is its circumradius and $O$ its circumcentre. The Pythagorean Theorem gives us $AO=\sqrt{6^2+8^2}=10$, so $OP=OQ=10$. Thus $P$ has $y$-coordinate 10 and $Q$ $y$-coordinate -10 and $PQ=20$. The altitude from $A$ to $\overline{PQ}$ is $6$ ($\overline{PQ}$ is on the $y$-axis, hence this altitude is just the $x$-coordinate of $A$), so $[\triangle APQ]=\dfrac{20\cdot6}{2}=60\implies\boxed{\textbf{(D)}\ 60}$.

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AMC 10 Problems and Solutions

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