2 points, diagonal line equation on a graph

[Guest]

Hello,

I am trying to help my daughter write the equation for a line given 2 end points on a graph, the line does not intersect y, and is diagonal. Points are (2,30) and (24,50). May have this wrong but I have the slope as 10/11. Need some hlp getting our equation for this line with end points.

Any help would be appreciated.

[Baltuilhe]

Good night!

Use the following:
[tex]m(x-x_0)=y-y_0[/tex]
Where m is the slope of the line.
[tex]m=\dfrac{y_1-y_0}{x_1-x_0}=\dfrac{50-30}{24-2}=\dfrac{20}{22}=\dfrac{10}{11}[/tex]

So, your slope is right!

Now:
[tex]\dfrac{10}{11}(x-2)=y-30[/tex] You could have used (24,50), too! (x0,y0) is a known point, just it!

Solving:
[tex]10(x-2)=11(y-30)\\10x-20=11y-330\\10x-20+330=11y\\10x+310=11y\\y=\dfrac{10}{11}x+\dfrac{310}{11}[/tex]

Hope to have helped!

[Guest]

linear function
y=ax+b
slpoe and intecepts

https://youtu.be/c6tVUxy9Sfw

[Guest]

https://youtu.be/V974YfDFSvo science video

[HallsofIvy]

As said above, any (non-vertical) line in an xy-coordinate system can be written as y= ax+ b for some numbers a and b.

The graph goes through (2,30) so y= 30 when x= 2 so 30= 2a+ b and through (24,50) so y= 50 when x= 24 so 50= 24a+ b. Solve the equations 2a+ b= 30 and 24a+ b= 50 for a and b.

Since in both equations, b has coefficient 1, we can eliminate it by subtracting one equation from the other:
22a= 20 so a= 20/22= 10/11. (That is exactly the calculation for "slope" Baltuilhe gives).

Since a= 10/11, 2a+ b= 20/11+ b= 30, b= 30- 20/11= (330- 20)/11= 310/11.

The line is given by y= (10/11)x+ 310/11. That could also be written as 11y= 10x+ 310 or 11y- 10x= 310.

[Guest]

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