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 $(3m+1)(m-3)=0$

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 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

As sol.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:

$8=-6*(\frac{1}{\frac{8-b}{6}})-b$

Flip the inverse:

$8=-6*(\frac{6}{8-b})-b$

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)

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 $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{D}$

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

Page

Toolbox

Search