2014 AMC 10A Problems/Problem 14

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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); label("$P$",(0,10),SW); label("$Q$",(0,-10),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 \[(3m+1)(m-3)=0\]

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 altitude 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 3

Like Solution 2 but solving directly for intercepts (b):

1. Solve for m using: $8=6m+b$

\[m=\frac{8-b}{6}\]

2. Substitute into the other equation:

\[8=-6\cdot(\frac{1}{\frac{8-b}{6}})-b\]

Flip the inverse:

\[8=-6\cdot(\frac{6}{8-b})-b\]

Multiply $6$'s:

\[8=-(\frac{36}{8-b})-b\]


3. Multiply through by $8-b$ (Watch distributing minus!)

\[64-8b=-36-8b+b^2\]

4. Add $36$ to both sides, and cancel $-8b$ by adding to both sides:

\[100=b^2\]

$b=10$ (or $-10$)

The rest is as above.


Solution 4(Heron's Formula)

Since their sum is $0$, let the y intercepts be P$(0,a)$ and Q$(0,-a)$. The slope of $AP$ is $\frac{8-a}{6}$. The slope of AQ is $\frac{8+a}{6}$. Since multiplying the slopes of perpendicular lines yields a product of $-1$, we have $\frac{64-a^2}{36}=-1$, which results in $a^2=100$. We can use either the positive or negative solution because if we choose $10$, then the other y-intercept is $-10$; but if we choose $-10$, then the other y-intercept is $10$. For simplicity, we choose that $a=10$ in this solution.

Now we have a triangle APQ with points A$(6,8)$, P$(0,10)$, and Q$(0,-10)$. By the Pythagorean theorem, we have that $AP=\sqrt{6^2+2^2}=2\sqrt{10}$, and that $AQ=\sqrt{6^2+18^2}=6\sqrt{10}$. $PQ$ is obviously $10--10=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{D}$.

~smartninja2000

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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