Question

How do you write the equation of a line in slope intercept, point slope and standard form given (3,-2) and has the slope of 3/4?

Answer 1

`y=3/4x-17/4,y+2=3/4(x-3),3x-4y=17`

Explanation:

The equation of a line in `color(blue)"point-slope form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1)" a point on the line"`

`"here "m=3/4" and " (x_1,y_1)=(3,-2)`

substituting these values into the equation gives.

`y-(-2)=3/4(x-3)`

`rArry+2=3/4(x-3)larrcolor(red)" in point-slope form"`

The equation in `color(blue)"slope-intercept form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))`
where m represents the slope and b, the y-intercept.

distribute and simplify the point-slope equation into this form.

`rArry+2=3/4x-9/4`

`y=3/4x-9/4-2`

`rArry=3/4x-17/4larrcolor(red)" in slope-intercept form"`

The equation in `color(blue)"standard form"` is.

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

Rearrange the slope-intercept equation into this form.

`3/4x-y=17/4larr" multiply through by 4"`

`rArr3x-4y=17larrcolor(red)" in standard form"`