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

y-11=-7/2(x+3) is one answer and y+3=-7/2(x-1) is the other

Answer 2

7x+2y =1

Explanation:

Hello

use the points to find the slope (A)

`A=(-3-11)/(1-(-3)) \\A= (-14)/(4) \\\\\\A=-(7)/(2) \\\\\\`

with the slope and a point , let´s take J(-3,11)

`y-y_(0) = A (x-x_(0) )\\\\y-11=-(7)/(2) (x+3)\\y-11=-(7x)/(2) -(21)/(2) \\y+(7x)/(2) =-(21)/(2) +11\\\\y+(7x)/(2) =0.5\\\\\\`

must be multiplied by two to eliminate the fractional and then order

`2*(y+(7x)/(2)) =2*0.5\\7x+2y =1`