AMC 10 Step by Step

2017 AMC 10B Problem 10

P1-10~3 minPrint

The lines with equations ax−2y=cax-2y=c and 2x+by=−c2x+by=-c are perpendicular and intersect at (1,−5)(1, -5) . What is cc ?

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Key insight

The slopes are a/2 and -2/b, and perpendicular slopes multiply to -1, so a = b. Putting (1, -5) into both equations then gives c.

Solution

There are three unknowns, aa, bb and cc, and the problem gives three facts: the lines are perpendicular, and the point (1,−5)(1, -5) is on each line. Turn each fact into an equation.

1. Slope of the first line. Solve ax−2y=cax - 2y = c for yy: y=a2x−c2y = \frac{a}{2}x - \frac{c}{2}. The slope is a2\frac{a}{2}.
2. Slope of the second line. Solve 2x+by=−c2x + by = -c for yy: y=−2bx−cby = -\frac{2}{b}x - \frac{c}{b}. The slope is −2b-\frac{2}{b}.
3. Perpendicular. Perpendicular lines have slopes that multiply to −1-1:
a2⋅(−2b)=−ab=−1,\frac{a}{2} \cdot \left(-\frac{2}{b}\right) = -\frac{a}{b} = -1,
so a=ba = b.
4. The point is on the first line. Put x=1x = 1, y=−5y = -5: a+10=ca + 10 = c.
5. The point is on the second line. Put x=1x = 1, y=−5y = -5: 2−5b=−c2 - 5b = -c, so c=5b−2c = 5b - 2.
6. Since a=ba = b, the last two equations say a+10=5a−2a + 10 = 5a - 2. So 4a=124a = 12 and a=3a = 3.
7. Then c=3+10=13c = 3 + 10 = 13.
8. Check: the lines are 3x−2y=133x - 2y = 13 and 2x+3y=−132x + 3y = -13. Both contain (1,−5)(1, -5): 3+10=133 + 10 = 13 and 2−15=−132 - 15 = -13. Their slopes 32\frac32 and −23-\frac23 multiply to −1-1.

The answer is (E) 13\boxed{\textbf{(E)}\ 13}.

Why this works

Count the unknowns and count the facts. Three unknowns need three equations, and here each sentence of the problem gives one. Write every fact as an equation before solving; the perpendicular condition is the simplest one and cuts the work down right away.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Slopes first. The line ax−2y=cax - 2y = c has slope a2\frac{a}{2}, and 2x+by=−c2x + by = -c has slope −2b-\frac{2}{b}. Perpendicular lines have slopes whose product is −1-1:
a2⋅(−2b)=−ab=−1⟹a=b.\frac{a}{2} \cdot \left(-\frac{2}{b}\right) = -\frac{a}{b} = -1 \quad\Longrightarrow\quad a = b.

Now the common point (1,−5)(1, -5) lies on both lines:
a(1)−2(−5)=c⇒c=a+10,2(1)+b(−5)=−c⇒c=5b−2.a(1) - 2(-5) = c \Rightarrow c = a + 10, \qquad 2(1) + b(-5) = -c \Rightarrow c = 5b - 2.
With a=ba = b, set a+10=5a−2a + 10 = 5a - 2, so 4a=124a = 12 and a=3a = 3. Then c=3+10=13c = 3 + 10 = 13.

Check: the lines are 3x−2y=133x - 2y = 13 and 2x+3y=−132x + 3y = -13; both pass through (1,−5)(1, -5) and their slopes 32\frac32 and −23-\frac23 multiply to −1-1.

The answer is (E) 13\boxed{\textbf{(E)}\ 13}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.