#hi how do you find the equation of l1?
frigid brook
spring sinewBOT
frigid brook
<@&286206848099549185>
frigid brook
<@&286206848099549185>
tall beacon
It's ez if you've done a-c
Oh wait actually slightly less ez than I thought
Anyways slope of l1 is negative reciprocal of line l2 cus perpendicular
So for example if slope of l2 is 3 then slope of l1 is -1/3
Now hmm we need to find the coordinates of P or B or any point in the line
But P isn't too bad
You know the coordinates of A and the coordinates of Q
So you can do a 90 degree rotation counterclockwise centered around A to get the coordinates for P
Since they're both on the circle centered around a
Now my dumb butt doesn't remember how to do that buuuuuuut
My smart butt knows how to derive it
so we first treat the coordinates of point A as a complex number w = a + bi and point Q as v = c +di
We need to shift A to be the origin, so we calculate (v - w) to be Q shifted to A being as the origin
then to rotate 90 degrees counterclockwise we multiply (v - w) by e^(i*pi/2)
Finally we need to translate the point back to O being the origin so we add w back in
And point P will be at the point represented by (v-w)e^(i pi/2)+ w
Hopefully you know about complex numbers cus if not
Uhh I'll have to get a piece of paper to do those calculations myself and tell you what a 90 degree counterclockwise rotation is lmao
Which I don't wanna do
So yeah now you have point P and the slope so just write the equation of the line like you always do when you have a point on the line and its slope
Actually hold on I'll you tell you where to start - if you rotate (x, y) by 90 degrees counterclockwise about the origin then the endpoint is (-y, x)
I can do that much in my head
OK actually I can do all of it in my head
A = (a, b)
Q = (c, d)
QA= <c-a, d-b>
90 degree rotation counterclockwise vector is <-(d-b), c - a>
Point P is at (b-d+a, c-a+b)
@frigid brook i believe in u u can do the rest on ur own
I'm assuming you did a-C correctly 🙃
frigid brook
then to rotate 90 degrees counterclockwise we multiply (v - w) by e^(i*pi/2)
we have not learnt that we not supposed to know that yet ðŸ˜
tall beacon
eh I did those calculations for u
frigid brook
is there another way to get point P
tall beacon
hm
frigid brook
idk we haven’t learnt that yet tho
tall beacon
well you know P has to be on the circle and on line l1 
And we have the equation for the circle
tall beacon
So from the circle equation you can solve for the x coordinate of P in terms of the Y coordinate or vice versa but I'm not sure how to solve for a numerical value of the coordinate given that you don't know the other line's equation. Even if we knew that we're only tryna find point P to find the line equation to begin with
frigid brook
yeah im just as confuse d
tall beacon
u sure u don't at least know about rotation and such?
Cus we don't need complex numbers
That's just how I re-derive it
Oh wait hm
OK I got it now @frigid brook
we should find pou t B instead of P
The intersection of l1 and l2
Let B = (x1, y1)
From the equation for l2 we've already derived we have y1 = mx1 + b (I dunno what m and b are but you found those earlier)
From the circle we can solve for y1 in terms of x1 in 2 different ways
That'll give us two solutions, pick the one with larger y1 value based in the graph I'm guessing
And then plug it back into the l2 equation and solve for x1
Then l1 will be y = -1/m(x-x1)+y1
Ahhh wait hold on
I'm dumb B isn't on the circle
But we can find point B another way I guess
If we make the line from A to B
But finding the slope of that would also require trig or at least rotations which I'm guessing you don't know about yet either
OK I have it for real this time
Let A =(a, b) (again you have the exact coordinates
P = (x1, y1) is our unknown this time
make the line l4 going through A given by y =m(x - a) +b, where m is the slope of line l2
Then this will intersect in two places with the circle
The solution with the larger y1 value will be it
yeah that should work