Question

What is the equation of the following line written in general form?

Answer 1

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

$x-y+1=0$

Answer 2

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