Respuesta :

eq. of line through (1,2) and (4,5) is
[tex]y-y1= \frac{y2-y1}{x2-x1} (x-x1), y-2= \frac{5-2}{4-1} (x-1)[/tex][tex]y-2= \frac{3}{3} (x-1) y-2=x-1 y-2-x+1=0 y-x-1=0[/tex]
x-y+1=0

Answer:

[tex]x-y+1=0[/tex]

Step-by-step explanation:

To find the equation of the line , use the points from the graph of the line

(1,2) and (4,5)

First frame equation of the line [tex]y=mx+b[/tex]

[tex]slope (m)=\frac{y_2-y_1}{x_2-x_1} =\frac{5-2}{4-1} =1[/tex]

m=1, (1,2)

[tex]y=mx+b[/tex]

[tex]2=1(1)+b[/tex]

b=1

So equation of the line is

[tex]y=1x+1[/tex]

General form is Ax+By+C=0

Subtract 1x from both sides

[tex]y=1x+1[/tex]

[tex]-1x+y=1[/tex], subtract 1 from both sides

[tex]-1x+y-1=0[/tex]

Divide whole equation by -1

[tex]x-y+1=0[/tex]

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