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Difference between revisions of "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 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