I was very confused when I saw the thumbnail and saw a negative sign of the constant term, but I then noticed the equation, the constant is on the right hand side. Where I'm from, we use ax+by+c=0 rather than ax+by=c.
I've never used this formula in my math up to Calculus 2, the closest I've ever done is find the the least distence between a point and an equation using the derivative and distence formula of sqrt((x2-x1)²+(y2-y1)²) and then finding the minimum with Calculus
@@RobaB.-zh4eewhat i would do: Let (p,q) be the given point. Let ax+by+c=0 be the given line. Let a point (x,y) lie on the given line. Then, r = sqrt( (x-p)² + (y-q)² ) Differentiating w.r.t. x, (I will use r', and y' to mean dr/dx and dy/dx) r' = (2x-2p+(2y-2q)y')/2(sqrt((x-p)²+(y-q)²) y' = -a/b (from eqn of given line) Set r' equal to zero, and solve for x and y, substituting from eqn of given line as necessary. Then find value of x and y and find the corresponding value of r
In computer programming you can substitute the term √ ( a² + b²) into : a/ sin ( atan( a / b)) Because square root is considered heavy computation. If ax+by+c=0 is the general equation of a straight line, y= -ax/b so, gradient m=-a/b since m=y/x then a = y , b=x √ (a² + b²) is just hypotenuse to the triangle |\ | \ | \ | \ -------- Let angle t is in between the base of the triangle and hypotenuse. Angle, t=atan (a/b) To find hypotenuse, H sin t = opposite Since a= the height of the triangle, a is also opposite to the angle t. H= a/sin t Substitute t, H= a/sin ( atan(a/b)) Thus, √ (a²+b²) = a/sin ( atan(a/b))
Uh... square root may be an expensive operation, but trigonometric functions and they're inverses are even _more_ expensive. Often the absolute distance matters less than relative distances, so it makes more sense to just skip the square root entirely and leave it squared.
I have a more general proof: let's say you have a plane in R3 ax+by+cz+d=0 and a point (i,j,k). The shortest distant from point to the plane is the normal vector (a,b,c) from the point to the plane that is (i,j,k)+(a,b,c)t, substitute it in to ax+by+cz+d=0 and solve for t, you will get (ai+bj+ck+d)/(a^2+b^2+c^2). Then, the length of the vector is just (ai+bj+ck+d)/sqrt(a^2+b^2+c^2) when you multiply t and the magnitude of (a,b,c). You can also extent this to higher dimension
@@nekothecat Another even more general approach: if you have a hyperplane: a1x1+a2x2+a3x3+...+anxn = b, then is a normal vector to the hyperplane. Normalize this normal and call it n. Then n•n^T will be the orthogonal projector onto the normal direction. Etc.
using coordinate transformations: left translation by c/a, followed by rotation matrix [[a, b]; [-b, a]]/(a^2+b^2). (both of these are isometries, of course) Then the line ax + by = c maps to the vertical line x = 0 (note that we transform equations via the inverse transformations on x and y, namely x -> x + c/a and (x, y) -> (ax + by, -bx + ay)/(a^2+b^2)), and (x1, y1) maps to (a*x1 + b*y1 - c, a*y1 - b*x1 + bc/a)/(a^2+b^2). So the distance is just the x-coordinate of the last point, (a*x1 + b*y1 - c)/(a^2+b^2).
Here is another way (assuming some other backround knowledge). Instead of using the fact that is normal to the line we will use the fact that points along the line. Then and the form a parallelogram and we can look at as the base. The area of the parallelogram is given by the absolute value if the 2x2 determinant having these vectors as columns. The distance we want is just the height of the parallelogram so take the area divided by the length of the base. Another way, embed the vectors in R3 by adding a 3rd component of 0 and use the cross product.
Interesting I didn't think about using a vector to prove this method. The only way I knew was to use the fact that the distance(line) between the point and the linear equation has a Perpendicular relation. And use the multiplication of two inclinations which has a Perpendicular relation is -1. ❤
That's a nice way to do that, especially if you have to prove the formula to students who don't know what calculus or vectors are. You may want to try with calculus too
@@Universo-ye5to Given a point P=(h,k) on the plane and the line y=mx+q, you want to find the minimum distance between P and the generic point R=(x, mx+q). In order to do that, you want to minimize the function D^2(x)=(h-x)^2+(k-mx-q)^2 (you don't need the square root because both functions have the same critical points). Since D^2(x) is a parabola and its quadratric term is (1+m^2)>0 , we know for sure that the function has just one critical point and that's a minimum. The rest is up to you, just equal the derivative to zero. You'll find both the parametric point R on the line that minimizes the distance in terms of the known quantities h,k,m,q AND the square root of the formula D^2(x_m) (where x_m is abscissa of the minumum). Finally if you want to obatain the formula for the generic point P=(x,y) and the cartesian line ax+by=c, just substitute m=-a/b, q=c/b, h=x and k=y.
@@Universo-ye5to Given a point P=(h,k) and a generic point on a line L=(x, mx+q) the squared distance between the 2 points is the function D^2(x)=(x-h)^2+(mx+q-k)^2 (for our purpose we don't need the root[D^2(x)] because both of the functions share the same critical points). Since D^2(x) is a parabola and its leading coefficient is always positive (1+m^2) we know that the curve has just one critical point and that's a global minimum. The rest is up to you
Alternative method: Let c1=ax1+by1, then ax+by=c1 is a line parallel to ax+by=c which passes through the point (x1,y1). The lines ax+by=c and ax+by=c1 intersects the x axis at x=-c/an and x=-c1/a respectively. Construct the perpendicular segment from (-c/a,0) to the line ax+by=c1, and call the acute angle either of the lines makes with the x axis p. Then |tan(p)|=|a/b| since it’s the absolute value of the slope, and |sin(p)|=|tan(p)|/sqrt(1+tan^2(p))=|a|/sqrt(a^2+b^2). The length of the perpendicular segment, which is also the distance from the point (x1,y1) to the line ax+by=c, is thus (|c-c1|/|a|)|sin(p)|=|ax1+by1-c|/sqrt(a^2+b^2).
You could do more geometry problems. Those are very useful in game programming. For example: Given a 2D line and a 2D circle, find the intersection points between the line and the circle.
An intuitive explanation: The expression ax + by measures a magnitude along the vector (a, b), and that magnitude is a distance scaled by the length of (a, b). Plugging in an arbitrary point (x, y) gives you a measurement, let's call it 'd', which must be divided by |(a, b)| to be a true distance: d / |(a, b)| The number 'c' in the equation ax + by = c is also such a measurement, so it must be divided by |(a, b)| to give a true distance as well: c / |(a, b)| We therefore reduce the problem to one dimension by measuring along (a, b) and simply subtracting the measurements to find the difference between them. The distance between an arbitrary point and a point satisfying that equation must therefore be |d - c| / |(a, b)|
Approximately 3.7632 for the question by Brilliant, my proof is finding the radius with the point P, (5). Then finding where the line y = -3x/4 + 7/4 intersects the graph of x^2 + y^2 = 25. Then using the positive intersection point as j hat and the negative intersection point as i hat. After that I put i hat and j hat into a 2x2 matrix then find that matrix’s determinant.
Signed distance is cool. It can be thought as that the line ax+by+c=0 has two distinct sides (which can be swapped by negation of coefficents). So the sign of distance is the determinant of which side of line point resides in. It's very useful in computer graphics and so on.
Solution to the last question Area =1/2 * b * (h1+h2) h1 = 7/5 , h2 =18/5 (concept from this video) r = 5 b/2 = 24/5 (Pythagoras theorem) Which gives the area as 24 sq. units
v is perpendicular to the line. coefficients a and b are a result of the slope of the line, which is -a/b. it goes (a) times down, since it's negative, and (b) times to the right. a vector that goes (a) times to the right and (b) times up will be perpendicular to the line, hence V = . you can imagine the line as a vector. u = and the vector v = . if you take the dot product u•v you'll get (ab-ab), which is 0. whenever dot product is 0, these vectors are orthogonal, in other words, perpendicular in R2
The vector proof is far simpler than an algebraic based solution where the slopes of the line and the perpendicular are related by -1. It will ultimately lead to the same result, but is a lot more work.
It says that the video was published 1 hour ago until I realized that your pinned comment was posted 19 hours ago, meaning that the video was actually posted more than 19 hours ago. RUclips resets the "Published" timestamp when the video was posted private or unlisted, set to public.
God sent His son Jesus to die for our sins on the cross. This was the ultimate expression of God's love for us. Then God raised Jesus from the dead on the third day. Please repent and turn to Jesus and receive Salvation now before it's too late. The end times written about in the Holy Bible are already happening in the world. Jesus loves you so soooo much ❤️ but time is almost up.
Can you do an equation where a is tetrated by x plus b exponentiated by x plus c multiplied by x plus d equal to 0 Can you make a formula for it too ? Pls I love your videos I can do calculus at age of 10 because of you
Divide the quadrilateral into 2 triangles, one triangle OAB and PAB. Now find the perpendicular distance of the point P and O from the line by using calculus, or direct formula (ax1+by1-c/root(a^2+b^2)), and add the areas of the 2 triangles which is just 1/2*base(AB is the common base for both)*height( starting from O for triangle OAB and from P for the triangle PAB). U will get ur answer. It is a bit lengthy, but an easy question.
Hmm...one can find the slope of any tangent line, etc, -1/m or something like that, can be computed easily, and then one can find the equation of the line using that passes through the point with that slope...I suppose one could have found what the equation of a perpendicular vector is instead, lol, using the dot product or something (using any two points of the line, one can find a vector that goes in the direction of the line, etc)...I mean, in order to use vectors at all here, lol...well, I suppose the way I would have normally done it sort of uses vectors, I would probably try to find the slope of the perpendicular using the dot product anyway, lol...
Not a fan of formulas though there are some exceptions with regard to difficult theorems. These are not formulas to memorise but you should be able to derive it with your knowledge of calculus.
P = (x₁, y₁) = e₁₂ + x₁e₂₀ + y₁e₀₁ L = (ax + by = c) = ae₁ + be₂ - ce₀ distance(P, L)/|P||L| = |P ∨ L|/|P||L = |*(*P ∧ *L)|/|P||L| = |*(-ce₀₁₂ + x₁ae₁₂₀ + y₁be₂₀₁)|/(√(a² + b²))(1) = |x₁a + y₁b - c|/√(a² + b²) Which looks like the solution the video got as well. It's nice that there's a simple mechanical algorithm for when you don't want to have to think about the problem.
The formula |P ∨ L/|P||L| from projective geometric algebra works in hyperbolic and elliptic geometries in addition to Euclidean ones, but still only works in these kinds of simple geometries.
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ans of final question: 48 square unit
I was very confused when I saw the thumbnail and saw a negative sign of the constant term, but I then noticed the equation, the constant is on the right hand side. Where I'm from, we use ax+by+c=0 rather than ax+by=c.
THANK YOU
Where are you from?? I'm from Mexico and here we also do it like ax+by+c=0
I've never used this formula in my math up to Calculus 2, the closest I've ever done is find the the least distence between a point and an equation using the derivative and distence formula of sqrt((x2-x1)²+(y2-y1)²) and then finding the minimum with Calculus
How only used Calculus get correct answer tell me?
@@RobaB.-zh4eewhat i would do:
Let (p,q) be the given point.
Let ax+by+c=0 be the given line.
Let a point (x,y) lie on the given line.
Then, r = sqrt( (x-p)² + (y-q)² )
Differentiating w.r.t. x, (I will use r', and y' to mean dr/dx and dy/dx)
r' = (2x-2p+(2y-2q)y')/2(sqrt((x-p)²+(y-q)²)
y' = -a/b (from eqn of given line)
Set r' equal to zero, and solve for x and y, substituting from eqn of given line as necessary. Then find value of x and y and find the corresponding value of r
have u taken any linear algebra courses?
You’d see this in calc 3
MUCH easier.
In computer programming you can substitute the term
√ ( a² + b²)
into :
a/ sin ( atan( a / b))
Because square root is considered heavy computation.
If
ax+by+c=0 is the general equation of a straight line,
y= -ax/b
so, gradient m=-a/b
since m=y/x
then a = y , b=x
√ (a² + b²) is just hypotenuse to the triangle
|\
| \
| \
| \
--------
Let angle t is in between the base of the triangle and hypotenuse.
Angle, t=atan (a/b)
To find hypotenuse,
H sin t = opposite
Since
a= the height of the triangle,
a is also opposite to the angle t.
H= a/sin t
Substitute t,
H= a/sin ( atan(a/b))
Thus,
√ (a²+b²) = a/sin ( atan(a/b))
Uh... square root may be an expensive operation, but trigonometric functions and they're inverses are even _more_ expensive. Often the absolute distance matters less than relative distances, so it makes more sense to just skip the square root entirely and leave it squared.
I have a more general proof: let's say you have a plane in R3 ax+by+cz+d=0 and a point (i,j,k). The shortest distant from point to the plane is the normal vector (a,b,c) from the point to the plane that is (i,j,k)+(a,b,c)t, substitute it in to ax+by+cz+d=0 and solve for t, you will get (ai+bj+ck+d)/(a^2+b^2+c^2). Then, the length of the vector is just (ai+bj+ck+d)/sqrt(a^2+b^2+c^2) when you multiply t and the magnitude of (a,b,c). You can also extent this to higher dimension
@@nekothecat Another even more general approach: if you have a hyperplane:
a1x1+a2x2+a3x3+...+anxn = b, then is a normal vector to the hyperplane. Normalize this normal and call it n. Then n•n^T will be the orthogonal projector onto the normal direction. Etc.
using coordinate transformations: left translation by c/a, followed by rotation matrix [[a, b]; [-b, a]]/(a^2+b^2). (both of these are isometries, of course) Then the line ax + by = c maps to the vertical line x = 0 (note that we transform equations via the inverse transformations on x and y, namely x -> x + c/a and (x, y) -> (ax + by, -bx + ay)/(a^2+b^2)), and (x1, y1) maps to (a*x1 + b*y1 - c, a*y1 - b*x1 + bc/a)/(a^2+b^2). So the distance is just the x-coordinate of the last point, (a*x1 + b*y1 - c)/(a^2+b^2).
Here is another way (assuming some other backround knowledge).
Instead of using the fact that is normal to the line we will use the fact that points along the line. Then and the form a parallelogram and we can look at as the base. The area of the parallelogram is given by the absolute value if the 2x2 determinant having these vectors as columns. The distance we want is just the height of the parallelogram so take the area divided by the length of the base.
Another way, embed the vectors in R3 by adding a 3rd component of 0 and use the cross product.
Interesting I didn't think about using a vector to prove this method. The only way I knew was to use the fact that the distance(line) between the point and the linear equation has a Perpendicular relation. And use the multiplication of two inclinations which has a Perpendicular relation is -1. ❤
I used
to do that demonstration, but it was tooooo large, so this helps a lot
That's a nice way to do that, especially if you have to prove the formula to students who don't know what calculus or vectors are. You may want to try with calculus too
@@Zonnymaka how you do it with calculus?
@@Universo-ye5to Given a point P=(h,k) on the plane and the line y=mx+q, you want to find the minimum distance between P and the generic point R=(x, mx+q). In order to do that, you want to minimize the function D^2(x)=(h-x)^2+(k-mx-q)^2 (you don't need the square root because both functions have the same critical points). Since D^2(x) is a parabola and its quadratric term is (1+m^2)>0 , we know for sure that the function has just one critical point and that's a minimum. The rest is up to you, just equal the derivative to zero. You'll find both the parametric point R on the line that minimizes the distance in terms of the known quantities h,k,m,q AND the square root of the formula D^2(x_m) (where x_m is abscissa of the minumum).
Finally if you want to obatain the formula for the generic point P=(x,y) and the cartesian line ax+by=c, just substitute m=-a/b, q=c/b, h=x and k=y.
@@Universo-ye5to Given a point P=(h,k) and a generic point on a line L=(x, mx+q) the squared distance between the 2 points is the function D^2(x)=(x-h)^2+(mx+q-k)^2 (for our purpose we don't need the root[D^2(x)] because both of the functions share the same critical points).
Since D^2(x) is a parabola and its leading coefficient is always positive (1+m^2) we know that the curve has just one critical point and that's a global minimum. The rest is up to you
Alternative method:
Let c1=ax1+by1, then ax+by=c1 is a line parallel to ax+by=c which passes through the point (x1,y1). The lines ax+by=c and ax+by=c1 intersects the x axis at x=-c/an and x=-c1/a respectively.
Construct the perpendicular segment from (-c/a,0) to the line ax+by=c1, and call the acute angle either of the lines makes with the x axis p. Then |tan(p)|=|a/b| since it’s the absolute value of the slope, and |sin(p)|=|tan(p)|/sqrt(1+tan^2(p))=|a|/sqrt(a^2+b^2). The length of the perpendicular segment, which is also the distance from the point (x1,y1) to the line ax+by=c, is thus (|c-c1|/|a|)|sin(p)|=|ax1+by1-c|/sqrt(a^2+b^2).
You could do more geometry problems. Those are very useful in game programming.
For example: Given a 2D line and a 2D circle, find the intersection points between the line and the circle.
Thanks ,because I was thinking about this formula and you did it.Fantastic!!!
An intuitive explanation:
The expression ax + by measures a magnitude along the vector (a, b), and that magnitude is a distance scaled by the length of (a, b). Plugging in an arbitrary point (x, y) gives you a measurement, let's call it 'd', which must be divided by |(a, b)| to be a true distance: d / |(a, b)|
The number 'c' in the equation ax + by = c is also such a measurement, so it must be divided by |(a, b)| to give a true distance as well: c / |(a, b)|
We therefore reduce the problem to one dimension by measuring along (a, b) and simply subtracting the measurements to find the difference between them. The distance between an arbitrary point and a point satisfying that equation must therefore be
|d - c| / |(a, b)|
If slope of a line ax^ 2 + bx + c = 0 is - a/ b then how to find a vector perpendicular to this line?
Approximately 3.7632 for the question by Brilliant, my proof is finding the radius with the point P, (5). Then finding where the line y = -3x/4 + 7/4 intersects the graph of x^2 + y^2 = 25. Then using the positive intersection point as j hat and the negative intersection point as i hat. After that I put i hat and j hat into a 2x2 matrix then find that matrix’s determinant.
Signed distance is cool. It can be thought as that the line ax+by+c=0 has two distinct sides (which can be swapped by negation of coefficents). So the sign of distance is the determinant of which side of line point resides in. It's very useful in computer graphics and so on.
Low-dimension least squares
Area(AOB) + Area(APB) = 24 m²; the radius of the circle is 5, and the line orthogonal to AB and also passing by P is y = 4/3 x;
So OP is also orthogonal to AB
Solution to the last question
Area =1/2 * b * (h1+h2)
h1 = 7/5 , h2 =18/5 (concept from this video)
r = 5
b/2 = 24/5 (Pythagoras theorem)
Which gives the area as 24 sq. units
true
I don't get it. Why vector of V = ?
Can anyone help me? Thanks in advances.
v is perpendicular to the line. coefficients a and b are a result of the slope of the line, which is -a/b. it goes (a) times down, since it's negative, and (b) times to the right. a vector that goes (a) times to the right and (b) times up will be perpendicular to the line, hence V = .
you can imagine the line as a vector. u = and the vector v = . if you take the dot product u•v you'll get (ab-ab), which is 0. whenever dot product is 0, these vectors are orthogonal, in other words, perpendicular in R2
Wondering if you could help me out. If you have three known points in 3 space, how do you find a point closest to the three known points?
The area is 24u^2
Why exclude horizontal and vertical lines (a=0 and b=0)?
They have a distance to the point too.
I remember doing this in my precalc class, my school's math classes are crazy haha
The vector proof is far simpler than an algebraic based solution where the slopes of the line and the perpendicular are related by -1. It will ultimately lead to the same result, but is a lot more work.
Thank you
Your pfp 💀
It says that the video was published 1 hour ago until I realized that your pinned comment was posted 19 hours ago, meaning that the video was actually posted more than 19 hours ago. RUclips resets the "Published" timestamp when the video was posted private or unlisted, set to public.
correct. i think the video was available to members first
Used this a lot last year lol! Never thought of this
God sent His son Jesus to die for our sins on the cross. This was the ultimate expression of God's love for us. Then God raised Jesus from the dead on the third day. Please repent and turn to Jesus and receive Salvation now before it's too late. The end times written about in the Holy Bible are already happening in the world. Jesus loves you so soooo much ❤️ but time is almost up.
@@L17_8 Yeah man sure, I'd do math for now though...
@@cdkw2 not you commenting from another account😂😂
@@lakshya4876 I know we have similar names but we are not the same. cdkw is a minecraft guild so maybe he used it as well lol
@@lakshya4876 I get we both have similar names but it's a coincidence lol.
Can you do an equation where a is tetrated by x plus b exponentiated by x plus c multiplied by x plus d equal to 0
Can you make a formula for it too ?
Pls
I love your videos I can do calculus at age of 10 because of you
(1/2)(7/5)(9.6)+(1/2)(18/5)(9.6)=24
Can please make a video how to solve Cubic equations (easy method
for the question at the end i found the equation for the circle and found intersections but im not sure how to find the area
Divide the quadrilateral into 2 triangles, one triangle OAB and PAB. Now find the perpendicular distance of the point P and O from the line by using calculus, or direct formula (ax1+by1-c/root(a^2+b^2)), and add the areas of the 2 triangles which is just 1/2*base(AB is the common base for both)*height( starting from O for triangle OAB and from P for the triangle PAB). U will get ur answer. It is a bit lengthy, but an easy question.
100 Derivative part 2 please 😅
Where is the question?
In the video
@@alphazero339 im talking about the question he said he'll ask in the end of the video
Hmm...one can find the slope of any tangent line, etc, -1/m or something like that, can be computed easily, and then one can find the equation of the line using that passes through the point with that slope...I suppose one could have found what the equation of a perpendicular vector is instead, lol, using the dot product or something (using any two points of the line, one can find a vector that goes in the direction of the line, etc)...I mean, in order to use vectors at all here, lol...well, I suppose the way I would have normally done it sort of uses vectors, I would probably try to find the slope of the perpendicular using the dot product anyway, lol...
Not a fan of formulas though there are some exceptions with regard to difficult theorems. These are not formulas to memorise but you should be able to derive it with your knowledge of calculus.
Amazing
Why the formula has the +c intead of the -c?
See @saathvikbogam 's comment
Video used ax+by=c instead of the usual ax+by+c=0
Thx
10:56 shwo
scalar projection
🤩👍 thank you
Come to india this is the first thing taught in coordination geometry in india for high school students 🗿🗿
why do i know it the same but with +c?
SOLVE THE ROTATING ON A SLANT QUESTION FROM EARLY TRANSCEDENTALS 6TH EDITION
P = (x₁, y₁) = e₁₂ + x₁e₂₀ + y₁e₀₁
L = (ax + by = c) = ae₁ + be₂ - ce₀
distance(P, L)/|P||L| = |P ∨ L|/|P||L = |*(*P ∧ *L)|/|P||L| = |*(-ce₀₁₂ + x₁ae₁₂₀ + y₁be₂₀₁)|/(√(a² + b²))(1) = |x₁a + y₁b - c|/√(a² + b²)
Which looks like the solution the video got as well. It's nice that there's a simple mechanical algorithm for when you don't want to have to think about the problem.
@@angeldude101 It is proved by Lapo.
the main cocept in Deming regression
Thankfully not another video on quadratic
Here sir.
❤
do a calculation on bitcoin
Only valid in Euclidean spacrs
The formula |P ∨ L/|P||L| from projective geometric algebra works in hyperbolic and elliptic geometries in addition to Euclidean ones, but still only works in these kinds of simple geometries.
24
7
I give u a challenge
Prove L Hôpital rule using Lagrange Mean value theorem 🗿
What the heck?
Everyone is fine until you start to include non-latin letters in your algebraic formulas 😂
the answer is 25
This is the worst way to calculate the distance 🌵🌵🌵🌵🌵🌵🌵🌵🌵🌵🌵🌵🌵🌵of a point to a strait line 🌵
I hate geometry man :(
First!
God sent His son Jesus to die for our sins on the cross. This was the ultimate expression of God's love for us. Then God raised Jesus from the dead on the third day. Please repent and turn to Jesus and receive Salvation now before it is too late. The end times written about in the Holy Bible are already happening in the world. Jesus loves you so soooo much ❤️ but time is almost up.
&.. what does that have to do with the video?
Dude wtf does this have to do with a distance between a dot and a line?
And then they wonder why more and more people dislike Christians every day. Pushing their shit down our throats when it's not even relevant
So true!! Jesus is the only way into heaven!