Answer:
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Parallel lines have equal slopes.
Find the slope of the first line, mn:
Find the slope of the second line, ab:
Compare the slopes and find the value of k:
Answer:
[tex]k=\dfrac{3}{2}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4.3 cm}\underline{Slope formula}\\\\$m=\dfrac{y_2-y_1}{x_2-x_1}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ \\ are points on the line.\end{minipage}}[/tex]
Segment MN
[tex]\implies \textsf{slope}=\dfrac{y_N-y_M}{x_N-x_M}=\dfrac{1-0}{5-0}=\dfrac{1}{5}[/tex]
Segment AB
[tex]\implies \textsf{slope}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{k-2 \frac{1}{4}}{-2\frac{1}{4}-1\frac{1}{2}}=\dfrac{k-2 \frac{1}{4}}{-2\frac{1}{4}-1\frac{2}{4}}=\dfrac{k-2 \frac{1}{4}}{-3\frac{3}{4}}[/tex]
Parallel lines have the same slope.
Therefore, equate the found slopes for the two segments and solve for k:
[tex]\implies \dfrac{k-2 \frac{1}{4}}{-3\frac{3}{4}}=\dfrac{1}{5}[/tex]
Cross multiply:
[tex]\implies 5 \left(k-2 \frac{1}{4}\right)=-3\frac{3}{4}[/tex]
[tex]\implies 5k-10 \frac{5}{4}=-3\frac{3}{4}[/tex]
[tex]\implies 5k-11 \frac{1}{4}=-3\frac{3}{4}[/tex]
Add 11 ¹/₄ to both sides:
[tex]\implies 5k-11 \frac{1}{4}+11 \frac{1}{4}=-3\frac{3}{4}+11 \frac{1}{4}[/tex]
[tex]\implies 5k=(-3+11)+\left(-\frac{3}{4}+\frac{1}{4}\right)[/tex]
[tex]\implies 5k=8-\frac{1}{2}[/tex]
[tex]\implies 5k=7\frac{1}{2}[/tex]
Divide both sides by 5:
[tex]\implies 5k \div 5=7\frac{1}{2} \div 5[/tex]
[tex]\implies 5k \div 5=\dfrac{15}{2} \div 5[/tex]
[tex]\implies k=\dfrac{3}{2}[/tex]
Therefore, the value of k for which the two segments are parallel is ³/₂.
