Need some help to write the equations of these lines

[Tak.Phy]

Hi everyone
So I have this homework and I need some help.
I did not know how to write the equations here in the topic so I thought the best solution is to upload a picture of the notebook page.

 

[SteamKing]

These are all straight lines. Don't you know the equation of a straight line?

 

[HallsofIvy]

Geometrically, a straight line is determined by two points. Any (non-vertical) line can be written y= ax+ b. Choose two points on each line. Replacing x and y with the x and y coordinates of those points gives you two equations to solve for a and b.

For example, if one line goes through (1, 3) and (3, 0) then 3= a(1)+ b and 0= a(3)+ b so we have the equations a+ b= 3 and 3a+ b= 0. Subtract the first equation from the second to get 2a= -3 so that a= -3/2. Then the first equation becomes a+ b= (-3/2)+ b= 3 so b= 3+ 3/2= 9/2. With a=-3/2 and b= 9/2, the equation is y= -(3/2)x+ 9/2 which could also be written 2y= -3x+ 9 or 3x+2y= 9.

(A vertical line can be written "x= constant".)

 

[Tak.Phy]

Thanks Hallsoflvy.

P.S: My cousin used my laptop and it seems that he posted this topic. I am very very sorry about this and it will never happen again.

 

[asdf12312]

Imho that's a little confusing. to find slope remember its m=\frac{y_2 - y_1}{x_2-x_1} where it doesn't matter which order you use for y₂ or y₁. So to use example of line with (1,3) and (3,0), you could do it:

m = \frac{0-3}{3-1} ⇔ \frac{3-0}{1-3} = -\frac{3}{2}

y = mx+b, we want to find b, we already have m and can use one of the points above for (x,y):
0 = (-\frac{3}{2})(3)+b → b = \frac{9}{2}