### The equation of the line through the points (4,2) and (-8, -2) is 3y =...

The equation of the line through the points (4,2) and (-8, -2) is 3y = PC+ q where p and q are constants. Find the value of p?

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Y-yi/yii-yi=x-xi/xii-xi
y-2/-4=x-4/-12
-4(x-4)=-12(y-2)
-4x-8+12y=0
recal equ =3y
-4x-8+9y=0
therefore p=-8.
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(4,2) and (-8,-2)
Using the coordinate:
(x1,y1)= (4,2)
(x2,y2)= (-8,-2)
= -2-2/-8-4
=-4/-12
=1/3
Using the equation of a straight line: y=mx+c
let's use one of the coordinates:
(4,2)
substituting in the equation above
y=2, x=4, m=1/3
2=1/3*4+c
2=4/3+c
Using 3 to multiply through the equation:
2*3=4/3*3+3*c
6=4+3c
collect like terms:
6-4=3c
2=3c
c=2/3
Substituting back into y=mx+c
y=1/3x+2/3
Multiply through by 3
3*y=1/3x*3+2/3*3
3y=x+2
Comparing 3y=px+q with 3y=x+2
we have that the value of p is 1
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