Question

How do you write the equation of a line in slope intercept, point slope and standard form given (2,-3) and (4,-1)?

Answer 1

The equation of the line in slope-intercept form is `y= x -5`
in point `(2,-3)` slope form is `y+3 = x-2`
and in standard form is `-x +y = -5`

Explanation:

The slope of the line passing through `(2,-3) and (4,-1)` is `m= (y_2-y_1)/(x_2-x_1)= (-1+3)/(4-2)=2/2=1`

Let the equation of the line in slope-intercept form be `y=mx+c or y=x+c` The point (2,-3) will satisfy the equation . So, `-3= 2+c or c= -5`

Hence the equation of the line in slope-intercept form is `y= x -5`

The equation of the line in point `(2,-3)` slope form is `y-y_1=m(x-x_1) or y - (-3) = 1*(x-2) or y+3 = x-2`

The equation of the line in standard form `(ax+by=c)` is #y= x-5 or -x +y = -5# {Ans]