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Difference between revisions of "2019 AMC 10B Problems/Problem 4"

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==Solution 1==
 
==Solution 1==
  
−
If all lines satisfy the equation, then we can just plug in values for a, b, and c that form an arithmetic progression. Let's do a=1, b=2, c=3 and a=1, b=3, and c=5. Then the two lines we get are: <cmath>x+2y=3</cmath> <cmath>x+3y=5</cmath>
+
If all lines satisfy the condition, then we can just plug in values for <math>a</math>, <math>b</math>, and <math>c</math> that form an arithmetic progression. Let's use <math>a=1</math>, <math>b=2</math>, <math>c=3</math>, and <math>a=1</math>, <math>b=3</math>, <math>c=5</math>. Then the two lines we get are: <cmath>x+2y=3</cmath> <cmath>x+3y=5</cmath>
−
Use elimination: <cmath>y = 2</cmath> Plug this into one of the previous lines. <cmath>x+4 = 3 \Rightarrow x=-1</cmath> Thus the common point is <math>\boxed{A) (-1,2)}</math>
+
Use elimination to deduce <cmath>y = 2</cmath> and plug this into one of the previous lines. We get <cmath>x+4 = 3 \Rightarrow x=-1</cmath> Thus the common point is <math>\boxed{\textbf{(A) } (-1,2)}</math>
  
 
==Solution 2==
 
==Solution 2==
−
We know that <math>a,b,c</math> are an arithmetic progression, so if the common difference is <math>d</math> we can say <math>a,b,c = a, a+d, a+2d.</math> Now we have <math>ax+ (a+d)y = a+2d</math>, and expanding gives <math>ax + ay + dy = a + 2d.</math> Factoring gives <math>a(x+y-1)+d(y-2) = 0</math>. Since this must always be true, we know that <math>x+y-1 = 0</math> and <math>y-2 = 0</math>, so <math>x,y = -1, 2,</math> and the common point is <math>\boxed{A) (-1,2)}</math>.
+
We know that <math>a</math>, <math>b</math>, and <math>c</math> form an arithmetic progression, so if the common difference is <math>d</math>, we can say <math>a,b,c = a, a+d, a+2d.</math> Now we have <math>ax+ (a+d)y = a+2d</math>, and expanding gives <math>ax + ay + dy = a + 2d.</math> Factoring gives <math>a(x+y-1)+d(y-2) = 0</math>. Since this must always be true (regardless of the values of <math>x</math> and <math>y</math>), we must have <math>x+y-1 = 0</math> and <math>y-2 = 0</math>, so <math>x,y = -1, 2,</math> and the common point is <math>\boxed{\textbf{(A) } (-1,2)}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:07, 17 February 2019

Problem

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?

$\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

Solution 1

If all lines satisfy the condition, then we can just plug in values for $a$, $b$, and $c$ that form an arithmetic progression. Let's use $a=1$, $b=2$, $c=3$, and $a=1$, $b=3$, $c=5$. Then the two lines we get are: \[x+2y=3\] \[x+3y=5\] Use elimination to deduce \[y = 2\] and plug this into one of the previous lines. We get \[x+4 = 3 \Rightarrow x=-1\] Thus the common point is $\boxed{\textbf{(A) } (-1,2)}$

Solution 2

We know that $a$, $b$, and $c$ form an arithmetic progression, so if the common difference is $d$, we can say $a,b,c = a, a+d, a+2d.$ Now we have $ax+ (a+d)y = a+2d$, and expanding gives $ax + ay + dy = a + 2d.$ Factoring gives $a(x+y-1)+d(y-2) = 0$. Since this must always be true (regardless of the values of $x$ and $y$), we must have $x+y-1 = 0$ and $y-2 = 0$, so $x,y = -1, 2,$ and the common point is $\boxed{\textbf{(A) } (-1,2)}$.

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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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