Question

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

Answer 1

7x + 2y= -1 this is the answer

Answer 2

The standard form of a linear equation: Ax + By = C.

The slope-point formula:

`y-y_1=m(x-x_1)\\\\m=(y_2-y_1)/(x_2-x_1)`

We have:

`J(-3,\ 11)\to x_1=-3,\ y_1=11\\K(1,\ -3)\to x_2=1,\ y_2=-3`

Substitute:

`m=(-3-11)/(1-(-3))=(-14)/(4)=-(7)/(2)\\\\y-11=-(7)/(2)(x-(-3))\\\\y-11=-(7)/(2)x-(21)/(2)\ \ \ \ |\cdot2\\\\2y-22=-7x-21\ \ \ |+22\\\\2y=-7x+1\ \ \ \ |+7x\\\\7x+2y=1`

Answer: JK: 7x + 2y = 1