Question

perpendicular bisector of the line (-3,0) (0,5)

Answer 1

find the midpoint ((x1-x2/2), (y1-y2)/2).
calculate the slope and invert the fraction and inverse the sign.
slope = 3, so -1/3 is the perpendicular slope.

Answer 2

First you would find the midpoint using the midpoint formula. (x1 + x2 / 2) , (y1 + y2 / 2), then find the slope, and then take the opposite recipricol and there is the perpendicular slope

Answer 3

The line's equation is
\[y=\frac{5}{3}x+5 \leftrightarrow \frac{5}{3}x-y=-5\]
Whose normal vector is
\[\left(\begin{matrix}\frac{5}{3} \\ -1\end{matrix}\right)\]

This vector directs a line such as
\[-x-\frac{5}{3}y+c=0\]

The midpoint of the two points is (-1,5;2,5)
so
\[1,5-\frac{5}{3}\times 2,5+c=0 \leftrightarrow c=-3.75\]
The equation is then

\[-x-\frac{5}{3}y-3.75=0\]

Answer 4

what??

Answer 5

his equation/procedure is correct. Its just a precalculus/calculus approach.

Answer 6

im in alg 2 tri LOL

Answer 7

Ummm, you know what a vector is Alfart?
Let's just say I found the equation of the line first with regular math, then with the line equation you can deduce the vector perpendicular to the line. With the vector perpendicular to the line, you can deduce the equation of the line directed by the vector, in other words a line perpendicular to your first line. Then it's just a matter of making it pass through the middle of your two points.

Answer 8

Im all confused now LOL

Answer 9

um, what don't you understand? Vectors or line equations?

Answer 10

vectors i never heard of them