Equation of A Line Through Two Points

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[sandanth]

I loved the moment when me and my friend figured out on why the formula of a straight line is (y-y1)=m(x-x1)

Basically it's the same formula where the gradient is found separately from the equation of the line, aswell as substituting the "b" of the equation with b=y1-mx1

[timeonly]

After getting slope m=(y₂-y₁)/(x₂-x₁), we can just plug in that expression for m into the point-slope form, and get:
y - y₁ = [(y₂-y₁)/(x₂-x₁)] (x-x₁) . It's fine as is, but of course you can move the y₁ to the RHS if you want to give y as a linear function of x.

[A way to see or remember the 2-point form of a straight line thru 2 given points is to make it an equality of two ratios, where you create two SIMILAR right triangles: one with the two given points (with vertical leg y₂-y₁, and horiz. leg, x₂-x₁ & hypotenuse along the line to be found), and a 2nd right triangle betw an arbitrary point (x, y) with one of the two original given points (x₁, y₁) (with vert. leg y-y₁, and horiz. leg x-x₁, & hypotenuse on the line to be found).
Equating the ratios of vert. legs to horz. legs gives the 2-point form.]

[copernic]

y = ax+b
Obviously we have a = (y2-y1)/(x2-x1)
To find b, let's find the point (0, y0):
(y2-y0)/(x2-0) = (y1-y0)/(x1-0)
=> y0 = (y1.x2-y2.x1)/(x2-x1)
and that's the value of b.
So, y = (y2-y1)/(x2-x1) * x + (y1.x2-y2.x1)/(x2-x1)

(if x1=0 or x2=0, the calculations are a bit different)

[makesalsanotwar]

Can't you use the determinant of a matrix to find out the equation? The first line is (x y 1) the second (x1 y1 1) and the third is (x2 y2 1)

[dorismukabe]

Find the equation of a straight line through (-1, -2)with a gradient of -2/3

[ElliotMatteyOfDoom]

Xisuma used a lot of Gold for the iBuy tunnel. He must really believe in his soul.

[angrybhalo]

Bruh you could've just used two point method


(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)

One line answer

[pwmiles]

Another way is to do it with homogeneous coordinates. The homogeneous coordinates of a point are (x, y, 1). The line equation
y = mx + b
can be written as a 3D dot product
(m, -1, b).(x, y, 1) = 0

The homogeneous line coordinates (so-called) are obtained from the vector product. Introducing a factor k
k(m, -1, b) = (x1, y1, 1) x (x2, y2, 1)
km = y1 - y2
-k = x2 - x1
kb = x1y2 - x2y1

Eliminate k
m = (y2 - y1) / (x2 - x1)
b = (x2y1 - x1y2) / (x2 - x1)