Difference between revisions of "2003 AMC 10B Problems/Problem 11"

Revision as of 19:24, 27 November 2015

A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?

$\textbf{(A) } 2 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 20$

Using the point-slope formula, the equation of each line is

$y-15=3(x-10) \longrightarrow y=3x-15$ $y-15=5(x-10) \longrightarrow y=5x-35$

Substitute in $y=0$ to find the $x$ -intercepts.

$0=3x-15\longrightarrow x=5$ $0=5x-35\longrightarrow x=7$ The difference between them is $7-5=\boxed{\textbf{(A) \ } 2}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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

@@ Line 9: / Line 9: @@
 Using the point-slope formula, the equation of each line is
-<cmath>y-15=3(x-10) \longrightarrow y=3x-15\\
+<cmath>y-15=3(x-10) \longrightarrow y=3x-15</cmath>
-y-15=5(x-10) \longrightarrow y=5x-35</cmath>
+<cmath>y-15=5(x-10) \longrightarrow y=5x-35</cmath>
 Substitute in <math>y=0</math> to find the <math>x</math>-intercepts.
-<cmath>0=3x-15\longrightarrow x=5\\
+<cmath>0=3x-15\longrightarrow x=5</cmath>
-=5x-35\longrightarrow x=7</cmath>
+<cmath>0=5x-35\longrightarrow x=7</cmath>
 The difference between them is <math>7-5=\boxed{\textbf{(A) \ } 2}</math>.