Point G is on line segment FH. Given FH = 4x, GH = x, and FG = 2x + 10,
2 answers:
Answer:
30 units
Step-by-step explanation:
If point G is on line segment FH, then point G lies between points F and H.
Segment Addition Postulate states that given 2 points F and H, a third point G lies on the line segment FH if and only if the distances between the points satisfy the equation FG + GH = FH.
By segment addition postulate,
Given
FH = 4x,
GH = x,
FG = 2x + 10,
then
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Answer:
#1) y = -4x + 1; #2) parallel; #3) y = -1/2x + 5/2; #4) y = 2x + 4; #5) Never
Step-by-step explanation:
#1) We want a line that passes through (1, -3) and is parallel to y = 2-4(x-1). First we simplify our equation using the distributive property:
y = 2-4(x-1) = 2-4(x) -4(-1) = 2-4x--4 = 2-4x + 4 = -4x + 6
Lines that are parallel have the same slope; this means we want a line through (1, -3) with a slope of -4. Using point-slope form,
#2) We must first write our second equation in slope-intercept form by isolating the y term:
2x+y=7
Subtract 2x from each side"
2x+y-2x = 7-2x
y = -2x+7
This means the slope is -2; the slopes are the same, so the lines are parallel.
#3) The slope of our given equation is 2. To be perpendicular, the second line must have a slope that is a negative reciprocal (flipped and opposite signs); this makes it -1/2. Using point-slope form,
#4) The slope of Main Street on the diagram, found by using rise/run, is 2. This means the bike path will also have a slope of 2 in order to be parallel. The park entrance is at (0, 4). This makes the equation y = 2x+4.
#5) Two lines with the same slope are always parallel. They are never perpendicular.