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grujan50
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:

[tex]y-8=-\frac{1}{2}(x+16)[/tex]

Step-by-step explanation:

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

[tex]y-y_1=m(x-x_1)[/tex]

Where,

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Here,

[tex]x_1=-16, y_1=8, x_2=4, y_2=-2[/tex]

Hence, the equation of the line would be,

[tex]y-8=\frac{-2-8}{4+16}(x+16)[/tex]

[tex]y-8=-\frac{10}{20}(x+16)[/tex]

[tex]y-8=-\frac{1}{2}(x+16)[/tex]

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