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

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

The slope-point formula:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have:

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

Substitute:

$m=\dfrac{-3-11}{1-(-3)}=\dfrac{-14}{4}=-\dfrac{7}{2}\\\\y-11=-\dfrac{7}{2}(x-(-3))\\\\y-11=-\dfrac{7}{2}x-\dfrac{21}{2}\ \ \ \ |\cdot2\\\\2y-22=-7x-21\ \ \ |+22\\\\2y=-7x+1\ \ \ \ |+7x\\\\7x+2y=1$

Answer: JK: 7x + 2y = 1