This is not a remainder theorem stuffs, but, please solve and explain to me this Problem. Find the equations of the lines containing the point(4,7) and passing at a distance 1 from the origin. Thanks
The two lines in slope-point form are: y - 7 = (12/5)(x - 4) and y - 7 = (4/3)(x - 4) The way I did it would take a lot of time to explain. Basically, if the lines pass a distance 1 from the origin, that means the lines are tangent to the circle x²+y²=1. Which means each line has one point on that unit circle (call this point (x*, y*)), and another point at (4, 7). So we know (x*, y*) satisfies y* - 7 = m(x* - 4) for some slope m. Also, since we know a circle's tangent lines are perpendicular to the diameter at (x*, y*) (and that perpendicular lines have negative reciprocal slopes), we also know that (x*, y*) satisfies y* - 0 = (-1/m)(x* - 0) for the same m. Solve each linear equation for m, and substitute one expression for m into the other equation, to get (y - 7)/(x - 4) = -x/y. Rearrange and use x²+y²=1 to get a new line: y = (-4/7)x + 1/7. The points on this line that intersect the unit circle are the 2nd points for the two original lines we wanted. Replace y with (-4/7)x + 1/7 in the equation x²+y²=1, and get a quadratic in x. The roots of this quadratic will be x=12/13 and x=-4/5. Plug these x-values back into y = (-4/7)x + 1/7 to get the corresponding y-values. The points are (x,y) = (12/13, -5/13) and (x,y) = (-4/5, 3/5). Finally, use these coordinates to find the slopes of the two lines: m=12/5, and m=4/3. There may be an easier way, but this is what I found.
![A-Level Maths: B6-11 [Polynomials: Introducing the Factor Theorem]](/img/n.gif)
I just want to see the kids reactions when he teaches them synthetic division
10:58 Choose easy things for yourself rather than hard things
Omg this was much needed.
I just learnt how to do this in my AP Math and I'm kinda struggling
With this video I'm sure I will ace my exams
Excellent sir G.
Outstanding.!!!
Good teacher
"Do we use Maths in Focus really?" 😀
2:28 "I am gonna put threes there everywhere i see Ex-es" 🤣
Why not discuss the sub factorial isn't that be great🤔
i was looking forward for the proof of it .... :(
This is not a remainder theorem stuffs, but, please solve and explain to me this Problem. Find the equations of the lines containing the point(4,7) and passing at a distance 1 from the origin. Thanks
The two lines in slope-point form are:
y - 7 = (12/5)(x - 4)
and
y - 7 = (4/3)(x - 4)
The way I did it would take a lot of time to explain. Basically, if the lines pass a distance 1 from the origin, that means the lines are tangent to the circle x²+y²=1. Which means each line has one point on that unit circle (call this point (x*, y*)), and another point at (4, 7). So we know (x*, y*) satisfies y* - 7 = m(x* - 4) for some slope m. Also, since we know a circle's tangent lines are perpendicular to the diameter at (x*, y*) (and that perpendicular lines have negative reciprocal slopes), we also know that (x*, y*) satisfies y* - 0 = (-1/m)(x* - 0) for the same m.
Solve each linear equation for m, and substitute one expression for m into the other equation, to get (y - 7)/(x - 4) = -x/y. Rearrange and use x²+y²=1 to get a new line: y = (-4/7)x + 1/7.
The points on this line that intersect the unit circle are the 2nd points for the two original lines we wanted. Replace y with (-4/7)x + 1/7 in the equation x²+y²=1, and get a quadratic in x. The roots of this quadratic will be x=12/13 and x=-4/5.
Plug these x-values back into y = (-4/7)x + 1/7 to get the corresponding y-values. The points are (x,y) = (12/13, -5/13) and (x,y) = (-4/5, 3/5).
Finally, use these coordinates to find the slopes of the two lines: m=12/5, and m=4/3.
There may be an easier way, but this is what I found.
Finally I got it. Thanks a lot for the help. Hopefully, I can approach you AGAIN if ever I need your awesome explanations. ; )
Hello SIR 👋
I think you should leave a reply for all your students here 💚🌸💯