How to solve (3,2) and (5,6) and turn it into 2x-y=4

You can start with the formula

... ∆y(x -x₁) -∆x(y -y₁) = 0

where ∆x = (x₂ -x₁) and ∆y = (y₂ -y₁). It does not matter which point you consider 1 or 2, but it is usually convenient to choose them so the differences are positive.

... (∆x, ∆y) = (5, 6) - (3, 2) = (2, 4)

For purposes of the above equation, it is convenient to divide these differences by any common factor they might have. Here, that factor is 2, so we can use

... (∆x, ∆y) = (1, 2) . . . . . . where (1, 2) = (2/2, 4/2)

Then, using ∆y=2, ∆x=1, x₁=3, y₁=2, the equation for the line becomes

... 2(x -3) - 1(y -2) = 0

... 2x -y -4 = 0 . . . . . . . simplified to general form

... 2x -y = 4 . . . . . . . . . add 4 to put in standard form.

The equation used here is a variation of the 2-point form of the equation for a line. That equation is usually seen as

... y -y₁ = (y₂ -y₁)/(x₂ -x₁)·(x -x₁)

In the terms used above, it can be written as ...

...  y -y₁ = ∆y/∆x·(x -x₁)

Multiplying by ∆x and subtracting the left side, we see it as above:

... ∆y·(x -x₁) - ∆x·(y -y₁) = 0

- (2, -7)

- lines r and s and lines t and u must be parallel

answer: answer is b.

[m < b = 120º]

step-by-step explanation: