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

Revision as of 18:03, 26 November 2011

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.

\[0=3x-15\longrightarrowx=5
0=5x-35\longrightarrowx=7\] (Error compiling LaTeX. Unknown error_msg)

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 16: / Line 16: @@
 <cmath>0=3x-15\longrightarrowx=5
 =5x-35\longrightarrowx=7</cmath>
-The difference between them is <math>7-5=\boxed{\mathrm{(A) \ } 2}</math>.
+The difference between them is <math>7-5=\boxed{\textbf{(A) \ } 2}</math>.
 ==See Also==
 {{AMC10 box|year=2003|ab=B|num-b=10|num-a=12}}