Question

question 2
a. on a coordinate plane (graph paper), create $\triangle lmn$ where $\overline{ln}$ is a vertical line segment and $mn$ is a horizontal line segment.
b. then create $\overline{qr}$ parallel to $\overline{lm}$ such that point $q$ is on $\overline{ln}$ and point $r$ is on $\overline{mn}$.
c. label points $l$, $m$, $n$, $q$, and $r$ with their coordinates.
d. use the slope formula to confirm that $\overline{qr}$ is parallel to $\overline{lm}$.

Explanation:

Step1: Assume Coordinates

Let's assume coordinates for points \( L \), \( M \), \( N \) to proceed. Let \( L=(x_1,y_1) \), \( M=(x_2,y_2) \), \( N=(x_3,y_3) \). Given \( LN \) is vertical, so \( x_1 = x_3 \) (vertical lines have same \( x \)-coordinate), and \( MN \) is horizontal, so \( y_2 = y_3 \) (horizontal lines have same \( y \)-coordinate). Let's pick specific values: Let \( L=(2,5) \), \( N=(2,2) \) (vertical, same \( x = 2 \)), \( M=(4,2) \) (horizontal with \( N \), same \( y = 2 \)).

Step2: Label Points Q, R

Point \( Q \) is on \( LN \), so \( x_Q = 2 \) (since \( LN \) is vertical \( x=2 \)). Let's take \( Q=(2,3) \). \( QR \) is parallel to \( LM \). First, find slope of \( LM \). Slope of \( LM \): \( m_{LM}=\frac{y_M - y_L}{x_M - x_L}=\frac{2 - 5}{4 - 2}=\frac{-3}{2} \). Since \( QR \) is parallel to \( LM \), slope of \( QR \) is also \( \frac{-3}{2} \). \( R \) is on \( MN \), so \( y_R = 2 \) (since \( MN \) is horizontal \( y=2 \)). Let \( R=(x_R,2) \). Slope of \( QR \): \( m_{QR}=\frac{y_R - y_Q}{x_R - x_Q}=\frac{2 - 3}{x_R - 2}=\frac{-1}{x_R - 2} \). Set equal to \( m_{LM} \): \( \frac{-1}{x_R - 2}=\frac{-3}{2} \). Cross - multiply: \( -2=-3(x_R - 2) \Rightarrow -2=-3x_R + 6 \Rightarrow 3x_R = 8 \Rightarrow x_R=\frac{8}{3}\approx2.67 \). So \( Q=(2,3) \), \( R=(\frac{8}{3},2) \), \( L=(2,5) \), \( M=(4,2) \), \( N=(2,2) \).

Step3: Verify Slopes

Slope of \( LM \): \( m_{LM}=\frac{2 - 5}{4 - 2}=\frac{-3}{2} \). Slope of \( QR \): \( m_{QR}=\frac{2 - 3}{\frac{8}{3}-2}=\frac{-1}{\frac{8 - 6}{3}}=\frac{-1}{\frac{2}{3}}=\frac{-3}{2} \). Since \( m_{LM}=m_{QR} \), \( QR\parallel LM \) (parallel lines have equal slopes).

Answer:

By choosing coordinates \( L=(2,5) \), \( M=(4,2) \), \( N=(2,2) \), \( Q=(2,3) \), \( R=(\frac{8}{3},2) \) and verifying slopes (both \( \frac{-3}{2} \)), we confirm \( QR\parallel LM \). The key was using slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) and showing equal slopes for parallel lines.