How do you find the general form of the line passing through (-1,2) and (2,5)?

First, find the slope. To do this, plug in values for this equation.
`m=(y2-y1)/(x2-x1)`
m is the slope and the values are your original coords.
`m=(5-2)/(2--1)`
`m=3/3`
`m=1`
Now that we have the slope, we use it to find the y-intercept, and the slope-intercept form.
We use point-slope for this.
`y-y1=m(x-x1)`
`y-2=1(x--1)`
`y-2=(x+1)`
`y-2=x+1`
`y=x+3`
The slope is 1, and the y-intercept is 3. The slope-intercept form is "y=x+3", and the point-slope form is "y-2=1(x+1)"

`"the equation of a line in "color(blue)"general form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By+C=0)color(white)(2/2)|)))`
where A is a positive integer and B, C are integers.

`"to begin express the equation in "color(blue)"slope-intercept form"`

`• y=mx+b`

`"where m represents the slope and b, the y-intercept"`

`"to calculate m use the "color(blue)"gradient formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))`
where `(x_1,y_1),(x_2,y_2)" are 2 coordinate points"`

`"the points are " (x_1,y_1)=(-1,2),(x_2,y_2)=(2,5)`

`rArrm=(5-2)/(2-(-1))=3/3=1`

`rArry=x+blarr" is the partial equation"`

`"to find b use either of the 2 given points"`

`"using " (2,5)" then"`

`5=2+brArrb=3`

`rArry=x+3larrcolor(red)" in slope-intercept form"`

`rArrx-y+3=0larrcolor(red)" in general form"`