Question 36327: (1) find the midpoint of the line segment with endpoints of (-3, -4) and (6-9)
Mid point = -3+6/2 and -4-9/2
= 3/2 and -13/2 are the midpoints
(2) find the slope of the line passing through (-3,-4) and (6,-9).
Slope of the line through the two point
slope = y2-y1 / x2-x1
= -9+4/6+3
= -5/9
(3) write the slope intercept form of the equation of the line through (-2,3) that is perpendicular to the line x+3y=4.
Found 2 solutions by ilana, kirubakaran:
Answer by ilana(307)   (Show Source):
You can put this solution on YOUR website! Oh, Algebra is fun!!
(1) For the midpoint, we use the formula ((x1+x2)/2,(y1+y2)/2), which is basically taking the averages of the coordinates. So here we have ((-3+6)/2,(-4-9)/2) = (3/2,-13/2).
(2) the slope is (change in y)/(change in x), or (y2-y1)/(x2-x1). So here we have (-9+4)/(6+3) = -5/9.
(3) First, we need to finsdd the slope of our line, which means we need the slope of x+3y=4. Writing this in slope-intercept form (y=mx+b) gives y=-x/3+4/3, which has a slope of -1/3. So, a line perpendicular to this has a slope of 3. Therefore, our line (through (-2,3) with slope 3) is y-3=3(x+2) (this is point-slope form), which in slope-intercept form is y=3x+9.
Answer by kirubakaran(5) (Show Source):
You can put this solution on YOUR website! (1) find the midpoint of the line segment with endpoints of (-3, -4) and (6-9)
Mid point = -3+6/2 and -4-9/2
= 3/2 and -13/2 are the midpoints
(2) find the slope of the line passing through (-3,-4) and (6,-9).
Slope of the line through the two point
slope = y2-y1 / x2-x1
= -9+4/6+3
= -5/9
(3) write the slope intercept form of the equation of the line through (-2,3) that is perpendicular to the line x+3y=4.
slope of x + 3y =4
3y = 4-x
y = 4/3-x/3
slope = m = -1/3
perpendicular slope = 3
therefore the equation is y-y1=m(x-x1)
Let (x1,y1)is (-2,3) y-3= =3(x+2)
y= 3x+9
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