Thank you so much !!! I spent 5 hours on Internet looking for an explanation of the formula for the distance of a point to a line ......and then....I found your video !!!! You made me happy !!!...excellent teacher , calm,clear..... and also smiling !!!
Derived it the hard way ... what a messed up beast, but in the end I got exactly the formula. Your derivation is much more elegant! I first constructed a perpendicular line b(x-x_0)-a(y-y_0)=0 through some point P=(x_0, y_0) for a general line ax+by+c=0 and then found the point of intersection (p,q) (horrible terms) and then used Pythagoras to find the distance between the two points d^2=(p-x_0)^2+(q-y_0)^2.
I couldn't find a proper derivation of this formula and it was really bothering me as it is needed in a lot of different topics . Thank you for the video , really appreciate it .
the gradient of the whole line is -a/b if you refer to the general formula of a gradient, it is (Y2-Y1)/(X2-X1), given that there are 2 points: (X1, Y1) and (X2, Y2). if you analyse this formula closely, you'll basically understand that the numerator (Y2-Y1) represents the y-distance and the denominator (X2-X1) represents the x-distance of the line gradient. In the context of this video, the y-distance equals to |-a| and the x-distance equals |b|. The modulus signs are important because, as Eddie mentioned, points could be found both under and above the line and we need to cancel out these variables with a fixed distance value. By applying this logic, TU=|b| which is the x-distance and UV=|-a|---->|a| because distance only has a positive value.
y=-a.x/b*-c/b , simply you said rise over run , but why x & c/b not counted in this case why they are neglected or ignored , why you only took -a/b but not he whole equation , please explain
y = -ax/b - c/b Mr Woo is only interested in the gradient of the line. The gradient is rise over run. But the gradient is also the coefficient of x in the equation y = mx + b so m is the gradient. m in this case is -a/b So rise is |a| and run is |b| he does not include -c/b part because that does not contribute to the gradient at all. that just shifts the graph up and down. it is the y-intercept.
How can the dimensions of triangle TUV be a and b they are already the coefficient of variable x and y of the horizontal line and we should use any other variable for the dimensions of TUV and while solving PQ/PR equation we can't cancel |b| with b of y cordinate they are different logically snd if this can happen then pls explain to me how?
***** Hi Mr Woo I'm 11 and I am in year 7 since you use to go to James ruse and you are an awesome mathematician I was wondering if you had any tips on getting 230/300 marks on the selective for me to get in James ruse. I am attending the selective test next year in year 8 so I can get into James ruse in year 9. So could you tell me any ways I could study or improve for the test and how much times I should study a day for it (anything that can improve me will be fine). And I also want to improve on my flaws last year when I did the test I didn't study for the selective test I missed a lot of questions I wasn't focusing and I was struggling And the worst part was I get a disadvantage because of my age. Hi Mr Woo I'm 11 and I am in year 7 since you use to go to James ruse and you are an awesome mathematician I was wondering if you had any tips on getting 230/300 marks on the selective for me to get in James ruse. I am attending the selective test next year in year 8 so I can get into James ruse in year 9. So could you tell me any ways I could study or improve for the test and how much times I should study a day for it (anything that can improve me will be fine). And I also want to improve on my flaws last year when I did the test I didn't study for the selective test I missed a lot of questions I wasn't focusing and I was struggling And the worst part was I get a disadvantage because of my age.
Thank you so much !!! I spent 5 hours on Internet looking for an explanation of the formula for the distance of a point to a line ......and then....I found your video !!!! You made me happy !!!...excellent teacher , calm,clear..... and also smiling !!!
Derived it the hard way ... what a messed up beast, but in the end I got exactly the formula. Your derivation is much more elegant! I first constructed a perpendicular line b(x-x_0)-a(y-y_0)=0 through some point P=(x_0, y_0) for a general line ax+by+c=0 and then found the point of intersection (p,q) (horrible terms) and then used Pythagoras to find the distance between the two points d^2=(p-x_0)^2+(q-y_0)^2.
I tried, but didn't work out... Maybe I mess up the arithmetic operation...
I decided to have a crack at this 'hard way' and realised it wasn't as bad as I thought it would be.
Let P = (x_0, y_0)
General line: ax + by + c = 0
Perpendicular line: bx - ay + c_0 = 0
We know P is on the perpendicular line
=> bx_0 - ay_0 + c_0 = 0
=> c_0 = ay_0 - bx_0
=> Perpendicular line: bx - ay + ay_0 - bx_0 = 0
The point of intersection (p,q) lies on both lines
ap + bq + c = 0
bp - aq + ay_0 - bx_0 = 0
ap + bq = - c
bp - aq = bx_0 - ay_0
p = (ac - b(bx_0 - ay_0))/(-a^2 - b^2) = (ac - b^2x_0 + aby_0)/(-(a^2+b^2)) = (b^2x_0 - aby_0 - ac)/(a^2+b^2)
q = (a(bx_0 - ay_0) + bc)/(-a^2 - b^2) = (abx_0 - a^2y_0 + bc)/(-(a^2 + b^2)) = (a^2y_0 - abx_0 - bc)/(a^2 + b^2)
p - x_0 = (b^2x_0 - aby_0 - ac)/(a^2+b^2) - x_0
= (b^2x_0 - aby_0 - ac)/(a^2+b^2) - x_0(a^2+b^2)/(a^2+b^2)
= (b^2x_0 - aby_0 - ac - x_0(a^2+b^2))/(a^2 + b^2)
= (b^2x_0 - aby_0 - ac - a^2x_0 - b^2x_0)/(a^2 + b^2)
= (-aby_0 - ac - a^2x_0)/(a^2+b^2)
= -a(by_0 + c + ax_0)/(a^2 + b^2)
= -a(ax_0 + by_0 + c)/(a^2 + b^2)
x_0 - p = a(ax_0 + by_0 + c)/(a^2 + b^2)
y_0 - q = y_0 - (a^2y_0 - abx_0 - bc)/(a^2 + b^2)
= y_0(a^2 + b^2)/(a^2 + b^2) - (a^2y_0 - abx_0 - bc)/(a^2 + b^2)
= (y_0(a^2 + b^2) - (a^2y_0 - abx_0 - bc))/(a^2 + b^2)
= (a^2y_0 + b^2y_0 - a^2y_0 + abx_0 + bc)/(a^2 + b^2)
= (b^2y_0 + abx_0 + bc)/(a^2 + b^2)
= b(by_0 + ax_0 + c)/(a^2 + b^2)
= b(ax_0 + by_0 + c)/(a^2+b^2)
d = sqrt((p - x0)^2 + (q - y_0)^2)
= sqrt((x_0 - p)^2 + (y_0 - p)^2)
= sqrt((a(ax_0 + by_0 + c)/(a^2 + b^2))^2 + (b(ax_0 + by_0 + c)/(a^2+b^2))^2)
= sqrt(a^2(ax_0 + by_0 + c)^2/(a^2 + b^2)^2 + b^2(ax_0 + by_0 + c)^2/(a^2 + b^2)^2)
= sqrt(((ax_0 + by_0 + c)^2/(a^2 + b^2)^2)(a^2 + b^2))
= sqrt((ax_0 + by_0 + c)^2/(a^2 + b^2))
= sqrt((ax_0 + by_0 + c)^2)/sqrt(a^2 + b^2) since (ax_0 + by_0 + c)^2 and (a^2 + b^2) are non-negative provided that a, b, c, x_0, and y_0 are real
= |ax_0 + by_0 + c|/sqrt(a^2 + b^2) since a, b, c, x_0, and y_0 are real
This guy is literally the god of maths
I couldn't find a proper derivation of this formula and it was really bothering me as it is needed in a lot of different topics . Thank you for the video , really appreciate it .
Eddie, you are a blessing packed into a 13 minute video
Thank you sir. Now i dont need to memorize the formula. More power to you!!
Good shit bro
I remember deriving this during my 8th grade exam (algebraicly) because I forgot 😂😂😂
An instructive proof, started in part 1. I like your involvement of the class, and your admission (7:47) that you do not YET know where Q is.
I really enjoy it.
I tried to work it out myself algebraically then I messed up... So I instantly searched Eddie Woo
Then after this video I literally became so happy
Thank you very much sir you really are the best math teacher i have ever seen my entire life
this is one of those math videos i smiled widely and became so happy after watching it because i just learned something good
Very clearly explained and such a simple solution.....ahhh similar triangles!
Great 👍🏻👍🏻👍🏻👍🏻
Can someone explain to me how he determined the measurements of the TU and UV lines? it was the only part i didn't understand ;-;
the gradient of the whole line is -a/b
if you refer to the general formula of a gradient, it is (Y2-Y1)/(X2-X1), given that there are 2 points: (X1, Y1) and (X2, Y2).
if you analyse this formula closely, you'll basically understand that the numerator (Y2-Y1) represents the y-distance and the denominator (X2-X1) represents the x-distance of the line gradient.
In the context of this video, the y-distance equals to |-a| and the x-distance equals |b|. The modulus signs are important because, as Eddie mentioned, points could be found both under and above the line and we need to cancel out these variables with a fixed distance value.
By applying this logic, TU=|b| which is the x-distance and UV=|-a|---->|a| because distance only has a positive value.
y=-a.x/b*-c/b , simply you said rise over run , but why x & c/b not counted in this case why they are neglected or ignored , why you only took -a/b but not he whole equation , please explain
y = -ax/b - c/b
Mr Woo is only interested in the gradient of the line. The gradient is rise over run. But the gradient is also the coefficient of x in the equation y = mx + b
so m is the gradient. m in this case is -a/b
So rise is |a| and run is |b|
he does not include -c/b part because that does not contribute to the gradient at all. that just shifts the graph up and down. it is the y-intercept.
thank you so much for this great explanation
Love it
How can the dimensions of triangle TUV be a and b they are already the coefficient of variable x and y of the horizontal line and we should use any other variable for the dimensions of TUV and while solving PQ/PR equation we can't cancel |b| with b of y cordinate they are different logically snd if this can happen then pls explain to me how?
THANK YOU SO MUCH✌✌
***** Hi Mr Woo I'm 11 and I am in year 7 since you use to go to James ruse and you are an awesome mathematician I was wondering if you had any tips on getting 230/300 marks on the selective for me to get in James ruse. I am attending the selective test next year in year 8 so I can get into James ruse in year 9. So could you tell me any ways I could study or improve for the test and how much times I should study a day for it (anything that can improve me will be fine). And I also want to improve on my flaws last year when I did the test I didn't study for the selective test I missed a lot of questions I wasn't focusing and I was struggling And the worst part was I get a disadvantage because of my age.
Hi Mr Woo I'm 11 and I am in year 7 since you use to go to James ruse and you are an awesome mathematician I was wondering if you had any tips on getting 230/300 marks on the selective for me to get in James ruse. I am attending the selective test next year in year 8 so I can get into James ruse in year 9. So could you tell me any ways I could study or improve for the test and how much times I should study a day for it (anything that can improve me will be fine). And I also want to improve on my flaws last year when I did the test I didn't study for the selective test I missed a lot of questions I wasn't focusing and I was struggling And the worst part was I get a disadvantage because of my age.
I can hear girls are ahead in his class.
great!!!
9:20
Omg that girl was so annoying
She was just asking a question
That’s rude