2014 AMC 10A Problems/Problem 14

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 $[\triangle APQ]=\dfrac{20\cdot6}{2}=60\implies\boxed{\textbf{(D)}\ 60}$ .

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2014 AMC 10A Problems/Problem 14

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