Question

Find an equation for the line in slope-intercept form
a)The line that passes through the origin and is parallel to the line containing (2,4) and (4,-4)
b)The line that passes through the point (1,7) and is perpendicular to the line x-3y+16=0

Answer 1

To find the equation of line we need two things
1)the point from where line is passes
2)slope of line somehow
a) line passes from origin(0,0)
given point (2,4) and (4,-4)
m=y2-y1
     x2-x1
m=-4-4    ⇒-8/2⇒-4
     4-2
equation of line is given by
m(x-x1)=(y-y1)
as x=y=0
-4(x)=y
y=-4x
b) x-3y+16=0
slope of line=1/3
slope  of perpendicular line will be=-1/slope of line=-3=m
the point from which line is passes (1,7)
equation of line
(x-x1)m=(y-y1)
(x-1)-3=(y-7)
-3x+3=y-7
-3x-y+3+7=0
-3x-y+10=0
3x+y-10=0

Answer 2

equation of a straight line in slope intercept form.
       y = m x + c

  slope = m =  slope of line passing thru two points =
             = (y2 - y1) / (x2- x1) = (-4 -4) /(4 -2 ) = -4
   the line passes thru origin,    so  substitute (0, 0) in  y = m x + c
         hence c  = 0.
so equation:    y = - 4 x      or  y + 4 x = 0
===================
b)
       slope of the line x - 3 y + 16 = 0      is   1/3
     slope of the line perpendicular to it:  -3    as product of slopes is -1.

       let the line be:    y = - 3 x + c
     it passes thru  (1, 7)  =>  7 = - 3 * 1 + c    => c = 10
         so the line is :  y = -3 x + 10  or,    y + 3x - 10 = 0