Need some help to write the equations of these lines

  • Thread starter Tak.Phy
  • Start date
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.


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


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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".)
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.
Imho that's a little confusing. to find slope remember its m=[itex]\frac{y_2 - y_1}{x_2-x_1}[/itex] where it doesnt 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 = [itex]\frac{0-3}{3-1}[/itex] ⇔ [itex]\frac{3-0}{1-3}[/itex] = -[itex]\frac{3}{2}[/itex]

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