Question

Write an equation of the line, in point-slope form, that passes through the two given points. points: (–16, 8), (4, –2)

Answer 1

This is the same question as from DINW 220, but I will answer.
This form of the straight line is an equation through the two points and it reads => y-y1= (y2-y1)/(x2-x1) (x-x1), point 1 (-16,8) and poin2 (4, -2) , x1= -16, y1= 8, x2= 4 and y2= -2 =>
y-8 = (-2-8)/(4-(-16)) (x-(-16)) => y-8= -10/(4+16) (x+16) =>
y-8= (-10/20) (x+16) If we simplify fraction (-10/20) we get (-1/2) =>
y-8=(-1/2) (x+16) we will multiply the both sides of the equation with number (2) we get 2y-16= ( -1) (x+16) => 2y-16= -x -16, we will add to the both sides number (+16) => 2y= -x than divide the both sides with number (2) we get y= (-1/2)x where -1/2 is coefficient of direction or (slope) and this linear function have not cut on the y axis, because it goes through a coordinate start ( 0,0) in the decartes coordinate system.

Answer 2

`y-8=-(1)/(2)(x+16)`

Explanation:

Since, the point-slope form of a line passes through
`(x_1, y_1)` and
`(x_2, y_2)` is,

`y-y_1=m(x-x_1)`

Where,

`m=(y_2-y_1)/(x_2-x_1)`

Here,

`x_1=-16, y_1=8, x_2=4, y_2=-2`

Hence, the equation of the line would be,

`y-8=(-2-8)/(4+16)(x+16)`

`y-8=-(10)/(20)(x+16)`

`y-8=-(1)/(2)(x+16)`