Question

one more problem i been stuck on
find equations of the lines that bisect the acute and obtuse angles formed by the graphs of 6x + 8y = -5 and 2x - 3y = 4

Answer 1

Take a point (u,v) and fine a relation between u and v so the distance from (u,v) to the two lines are the same

Answer 2

Square the two distances and make them equal, you get
\[
\frac{(6 u+8 v+5)^2}{6^2+8^2}=\frac{(2 u-3 v-4)^2}{2^2+3^2}

\]

Answer 3

Solve for v in terms of u

Answer 4

so wait how do i find the point between u and v?

Answer 5

Take any point (u,v). If it is on the bisector the distance from (u,v) to each of the line must be the same

Answer 6

so it is okay if I can use 2 and 1

Answer 7

You will get the two bisectors at once
\[
v=\left(18-5 \sqrt{13}\right) u+\frac{5 \sqrt{13}}{2}-10\\
v=\left(18+5 \sqrt{13}\right) u-\frac{5 \sqrt{13}}{2}-10

\]

Answer 8

No, (u,v) should be variables. Now exchanging u by x and v by y, you get
the equations of the two bissectors
\[
y=\left(18-5 \sqrt{13}\right) x+\frac{5 \sqrt{13}}{2}-10\\
y=\left(18+5 \sqrt{13}\right) x-\frac{5 \sqrt{13}}{2}-10

\]

Answer 9

Here is the graph of the lines and the bisectors

Answer 10

ohh wow thank you im taking my time to understand this concept

Answer 11

Notice that the product of the two slopes of the bisectors is =-1. Did we expect that?
\[
\left(18-5 \sqrt{13}\right) \left(18+5 \sqrt{13}\right)=-1

\]

Answer 12

The concept is easy, working out the solution is a bit tedious, but only requires the use of the quadratic formula.

Answer 13

ohh thank you very much without your help I would stay up really late just to get this figure out or worst I just skipped the problem

Answer 14

YW