Twenty percent (20%) of 90 equals A. 12 B. 15 C. 18 D. 21
1 answer:
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You have the correct answer. Nice work. If you need to see the steps, then see below
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First we need to find the midpoint of H and I
The x coordinates of the two points are -4 and 2. They add to -4+2 = -2 and then cut that in half to get -1
Do the same for the y coordinates: 2+4 = 6 which cuts in half to get 3
So the midpoint of H and I is (-1,3). The perpendicular bisector will go through this midpoint
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Now we must find the slope of segment HI
H = (-4,2) = (x1,y1)
I = (2,4) = (x2,y2)
m = (y2 - y1)/(x2 - x1)
m = (4 - 2)/(2 - (-4))
m = (4 - 2)/(2 + 4)
m = 2/6
m = 1/3
Flip the fraction to get 1/3 ---> 3/1 = 3
Then flip the sign: +3 ----> -3
So the slope of the perpendicular bisector is -3
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Use m = -3 which is the slope we found
and (x,y) = (-1,3), which is the midpoint found earlier
to get the following
y = mx+b
3 = -3*(-1)+b
3 = 3+b
3-3 = 3+b-3
0 = b
b = 0
So if m = -3 and b = 0, then y = mx+b turns into y = -3x+0 and it simplifies to y = -3x
So that confirms you have the right answer. I've also used GeoGebra to help confirm the answer (see attached)
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First we need to find the midpoint of H and I
The x coordinates of the two points are -4 and 2. They add to -4+2 = -2 and then cut that in half to get -1
Do the same for the y coordinates: 2+4 = 6 which cuts in half to get 3
So the midpoint of H and I is (-1,3). The perpendicular bisector will go through this midpoint
---------------------
Now we must find the slope of segment HI
H = (-4,2) = (x1,y1)
I = (2,4) = (x2,y2)
m = (y2 - y1)/(x2 - x1)
m = (4 - 2)/(2 - (-4))
m = (4 - 2)/(2 + 4)
m = 2/6
m = 1/3
Flip the fraction to get 1/3 ---> 3/1 = 3
Then flip the sign: +3 ----> -3
So the slope of the perpendicular bisector is -3
-----------------------
Use m = -3 which is the slope we found
and (x,y) = (-1,3), which is the midpoint found earlier
to get the following
y = mx+b
3 = -3*(-1)+b
3 = 3+b
3-3 = 3+b-3
0 = b
b = 0
So if m = -3 and b = 0, then y = mx+b turns into y = -3x+0 and it simplifies to y = -3x
So that confirms you have the right answer. I've also used GeoGebra to help confirm the answer (see attached)