finding the length of lines

[Clifford]

Can somebody check over these and see if I did them correctly? Thanks

\(\displaystyle \[
\begin{array}{l}
l_{lm} \; = \;\sqrt {\left( {a\; + \;b\; - \;0} \right)^2 \; + \;\left( {0\; - \;\left( {b\; + \;a} \right)} \right)^2 } \\
l_{lm} \; = \;\sqrt {\left( {a^2 \; + \;2ab\; + \;b^2 } \right)\; + \;\left( {a^2 \; + \;2ab\; + \;b^2 } \right)} \\
l_{lm} \; = \;\sqrt {2a^2 \; + \;4ab\; + \;2b^2 } \\
\end{array}
\]\)

and:

\(\displaystyle \[
\begin{array}{l}
l_{pn} \; = \;\sqrt {\left( { - b\; - \;\left( {\frac{{a\; + \;b}}{2}} \right)} \right)^2 \; + \;\left( { - a\; - \;\left( {\frac{{b\; + \;a}}{2}} \right)} \right)^2 } \\
l_{pn} \; = \;\sqrt {\left( {b^2 \; + \;\frac{{2ab\; + \;b^2 }}{2}\; + \;\frac{{a^2 \; + \;2ab\; + \;b^2 }}{4}} \right)\; + \;\left( {a^2 \; + \;\frac{{2ab\; + \;a^2 }}{2}\; + \;\frac{{b^2 \; + \;2ab\; + \;a^2 }}{4}} \right)} \\
l_{pn} \; = \;\sqrt {\frac{{4b^2 \; + \;4ab\; + \;2b^2 \; + \;a^2 \; + \;2ab\; + \;b^2 \; + \;4a^2 \; + \;4ab\; + 2a^2 \; + \;b^2 \; + \;2ab\; + \;a^2 }}{4}} \\
l_{pn} \, = \;\sqrt {\frac{{8b^2 \; + \;12ab\; + \;8a^2 }}{4}} \\
l_{pn} \, = \;\sqrt {2b^2 \; + \;3ab\; + \;2a^2 } \\
\end{array}
\]\)

 

[stapel]

What do "llm" and "lpn" stand for? What are the relationships between these and "a" and "b"? What were the original exercises?

Thank you.

Eliz.

 

[Clifford]

the first l stands for the length. lm is the line segment as well as pn. The line segments are points on a graph.

l(0, b + a)
m(a + b, 0)
p (a + b / 2, b + a / 2)
n ( -b, -a)

 

[Denis]

Yer sure good with LaTex, Cliff

 

[soroban]

Hello, Clifford!

You could make your algebra a bit easier . . .

\(\displaystyle L(0,\,a+b),\;\;M(a+b,\,0),\;\;P\left(\frac{a+b}{2},\,\frac{a+b}{2}\right),\;\;N(-b,\,-n)\)

Find: \(\displaystyle \,\overline{LM}\) and \(\displaystyle \overline{PN}\)

\(\displaystyle \overline{LM} \;= \;\sqrt{[(a+b)\,-\,0]^2\,+\,[0\,-\,(a+b)]^2} \;=\;\sqrt{(a+b)^2\,+\,(a+b)^2}\;=\;\sqrt{2(a+b)^2}\;=\;\sqrt{2}\,(a\,+\,b)\)


\(\displaystyle \overline{PN}\;=\;\sqrt{\left(\frac{a+b}{2}\,+\,b\right)^2\,+\,\left(\frac{a+b}{2}\,+\,a\right)^2} \;=\;\sqrt{\left(\frac{a+3b}{2}\right)^2\,+\,\left(\frac{3a+b}{2}\right)^2} \;=\;\sqrt{\frac{a^2\,+\,6ab\,+\,9b^2}{4}\,+\,\frac{9a^2\,+\,6ab\,+\,b^2}{4}}\)

. . \(\displaystyle = \;\sqrt{\frac{10a^2\,+\,12ab\,+\,10b^2}{4}} \;=\;\frac{\sqrt{10a^2\,+\,12ab\,+\,10b^2}}{2}\)

 

[Clifford]

Thanks for the response. A further question regarding these two lines is finding the area of a triangle. Those lines are the base and the height of the triangle. The area of the triangle is suppose to come out to (a+b)^2, but I don't see how plugging those two anwers into the bh/2 formula will give me this answer.