if
a and 3

2
are the slopes of perpendicular lines, a =
someone wanna explain negative recipricals to me?
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if
a and 3

2
are the slopes of perpendicular lines, a =
someone wanna explain negative recipricals to me?
you got it a/2 *3=1Quote:
Originally Posted by zoso
a=2/3
Malay
What do you mean a=? Are you just using it as another variable to the slope (m)? Usually the slopes of all lines are m=, regardless of them being perpendicular. In your case, the slope of the line perpendicular to a line with the slope of$\displaystyle 3/2$ would be $\displaystyle m=2/3$Quote:
Originally Posted by zoso
Negative reciprocals are basically just the opposite of the original fraction/whole number (just flip it).
Examples:
$\displaystyle m=3/1$, the negative reciprocal is: $\displaystyle 1/3$
$\displaystyle m=5/2$, the negative reciprocal is: $\displaystyle 2/5$
$\displaystyle m=1/5$, the negative reciprocal is: $\displaystyle 5$
$\displaystyle m=4/5$, the negative reciprocal is: $\displaystyle 5/4$
Remember, parallel lines have the same slope.
Hope this helps..
NineZeroFive
the question was if a line with the slope $\displaystyle \frac{a}{2}$ is perpendicular to a line with the slope $\displaystyle 3$ then what is the value of a?Quote:
Originally Posted by 905
which (this is just extending what Malay said) would mean that $\displaystyle 3$ and $\displaystyle \frac{a}{2}$ are negative reciprocal, which implies that multiplied together they equal 1, now solve for a...
$\displaystyle \frac{a}{2}\times 3=\neg 1$
$\displaystyle \frac{a}{2}=\frac{\neg 1}{3}$
$\displaystyle \boxed{a=\frac{\neg 2}{3}}$
check:
$\displaystyle \frac{a}{2}\times 3=\neg 1$
$\displaystyle \frac{\frac{\neg 2}{3}}{2}\times 3=\neg 1$
$\displaystyle \frac{\neg 2}{6}\times 3=\neg 1$
$\displaystyle \frac{\neg 6}{6}=\neg 1$
$\displaystyle \neg 1=\neg 1$
Interesting use of negative reciprocal:
Just a bit of info you might like nzf, in a linear equation like $\displaystyle y=mx+b$ to find the xintercept you multiply the yintercept by the slopes negative reciprocal.
~ $\displaystyle Q\!u\!i\!c\!k$
Slope are used to find family of lines.
You are given a line $\displaystyle ax+by+c=0$
Family of lines parallel to the given line
$\displaystyle ax+by+k=0$(Same coefficients of x and t means same slope, k is a parameter)
Family of lines perpendicular to the given line
$\displaystyle bxay+k=0$
or,$\displaystyle bx+ay+k=0$(Swapping the cofficients of x and y and multiplying anyone of then by 1 gives a perpendicular line, this is due to negative reciprocal)
Keep Smiling
Malay