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2014 AMC 10A Problems/Problem 14

Revision as of 11:10, 10 February 2014 by NeoMathematicalKid (talk | contribs) (→‎Solution 1: added labels to axes)

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 1

[asy]//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); 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); label("$a$", (0,5),W); label("$a$",(0,-5),W); label("$a$",(3,4),NW); // wanted to import graph and use xaxis/yaxis but w/e label("$x$",(9,0),E); label("$y$",(0,13),N); [/asy] Note that if the $y$-intercepts have a sum of $0$, the distance from the origin to each of the intercepts must be the same. Call this distance $a$. Since the $\angle PAQ = 90^\circ$, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\sqrt{6^2+8^2} = 10$, this means $a=10$, and the length of the hypotenuse is $2a = 20$. Since the $x$-coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}$.

Solution 2

We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$, and the sum of the y-intercepts is equal to 0, hence the $b, -b$.

Since both lines contain the point $(6,8)$, we can plug this into the two equations to obtain \[8=6m+b\] and \[8=-6\frac{1}{m}-b\]

Adding the two equations gives \[16=6m+\frac{-6}{m}\] Multiplying by $m$ gives \[16m=6m^2-6\] \[\implies 6m^2-16m-6=0\] \[\implies 3m^2-8m-3=0\] Factoring gives \[(6m+2)(m-3)=0\]

We can just let $m=3$, since the two values of $m$ do not affect our solution - one is the slope of one line and the other is the slope of the other line.

Plugging $m=3$ into one of our original equations, we obtain \[8=6(3)+b\] \[\implies b=8-6(3)=-10\]

Since $\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altutude to the hypotenuse is equal to the the x-coordinate of point $A$, or 6, the area of $\bigtriangleup APQ$ is equal to \[\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}\]

(Solution by bestwillcui1)

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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