Verified Answer :

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

Answer:

\(x-y+1=0\)

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 \(y=mx+b\)

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

m=1, (1,2)

\(y=mx+b\)

\(2=1(1)+b\)

b=1

So equation of the line is

\(y=1x+1\)

General form is Ax+By+C=0

Subtract 1x from both sides

\(y=1x+1\)

\(-1x+y=1\), subtract 1 from both sides

\(-1x+y-1=0\)

Divide whole equation by -1

\(x-y+1=0\)

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