Sunday, July 14, 2013

# What happens when a line intersects a bunch of parallel lines?

10:53 AM
You can expect questions in the GRE about a line that intersects a pair of parallel lines and in the process creates angle pairs. Let’s take a detailed look at them and understand the properties of these angle pairs. In the following figure, line  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} k_1$ intersects a pair of parallel lines $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} L_1$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} L_2$. Total of 8 angles are formed as a result of the intersection. The following figure shows these 8 angles.

The important thing to remember is that all angles shown in blue are equal to each other and all angles shown in red are equal to each other.
In addition you also get pairs of supplementary angles e.g. in the above figure a red angle and a  blue angle are supplementary to each other. In other words blue + red = 180º
Example:
In the above figure, if line  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} k$ intersects parallel lines  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} L_1$ and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} L_2$ , what is the value of $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} u+2x-a+y$?
A) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 0$
B) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 180$
C) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 360$
D) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x+2y$
E) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} 2y$
Solution:

Step 1: Angles $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x$and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} a$are corresponding angles and therefore $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x=a$. Plugging in the $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} a=x$ in the expression, we get:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin{align*} u+2x-a+y&=u+2x-x+y \\&=u+x+y\end{}
Step 2: Angles $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x$and $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} u$are supplementary angles and therefore $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x+u=180$. Substituting $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x+u=180$ in the expression we get:
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin{align*} u+2x-a+y &= u+x+y \\&=180+y\end{}
Since $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} u$ can not be negative, we can immediately rule out the first three options.
Step 3: However angles  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} y$ and  $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x$ are supplementary angles and therefore $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} x+y=180$.
\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \begin{align*} u+2x-a+y&=u+x+y \\&=180+y \\&=x+y+y \\&=x+2y \\\end{}