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# y=-2x+6 is the perpendicular bisector of the line segment

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CEO
Joined: 21 Jan 2007
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y=-2x+6 is the perpendicular bisector of the line segment [#permalink]  25 Sep 2007, 23:36
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?
Director
Joined: 11 Jun 2007
Posts: 932
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Kudos [?]: 72 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  25 Sep 2007, 23:42
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

don't you just plug it back into the new equation ?

1/2x - 1.5
1/2(3) - 1.5 = 0
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 354 [0], given: 4

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 00:05
beckee529 wrote:
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

don't you just plug it back into the new equation ?

1/2x - 1.5
1/2(3) - 1.5 = 0

argh my brain is not working at 4am...
Senior Manager
Joined: 13 Mar 2007
Posts: 295
Location: Russia, Moscow
Followers: 2

Kudos [?]: 22 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 04:41
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

perp bisector ==> neg reciprical
-2 ==> +1/2
Is this a rule???
VP
Joined: 10 Jun 2007
Posts: 1465
Followers: 6

Kudos [?]: 137 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 05:22
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

I get (-1,-2)
you know that the line bisects at x=3
plug this in the equation, you get y=-6+6 = 0
So the point is at (3,0)
This means point B must be:
(3-(7-3), 0-(2-0))
(-1, -2)
CEO
Joined: 29 Mar 2007
Posts: 2589
Followers: 16

Kudos [?]: 238 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 10:03
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

If -2 is the slope of the bisector than 1/2 is the slope of the line segment.

now we need to find the y intercept if we haveone. y=x/2+B use A's coordinates.

2=7/2+B B=-3/2 so now we have y=x/2-3/2

to find B:

u gotta find where the two lines intersect: x/2-3/2=-2x+6

-15/2=-5x/2 ---> 15/2=5x/2 ---> x=3, thus y=0.

Essentially 3,0 is the midpoint of line segment A,B. So since 7 is 4 away from 3, just subtract 3 by 4. We get -1 for x. and y use same logic. we get -2.

-1,-2.
Director
Joined: 09 Aug 2006
Posts: 765
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Kudos [?]: 72 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 10:25
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

Getting (-1, -2)

I drew out the figure. Happens that y=-2x+6 is downward sloping; intercepts the y axis at (0,6); intercepts the x axis at (3,0). Fortunately for us, it also happens that (3,0) is the point where the line y=-2x+6 bisects AB.

Once you figure this out its simple. Since one half of the line is 4 units away from the midpoint (3,0) on the x axis and 2 units up (on the y axis), the other half, to the left of the midpoint will also be 4 units away from the midpoint 3-4=-1 and 2 units down, -2. That gives us (-1, -2).
Senior Manager
Joined: 13 Mar 2007
Posts: 295
Location: Russia, Moscow
Followers: 2

Kudos [?]: 22 [0], given: 0

Re: MGMAT Geometry page 73 #9 [#permalink]  26 Sep 2007, 22:44
IrinaOK wrote:
bmwhype2 wrote:
y=-2x+6 is the perpendicular bisector of the line segment AB. If A has coordinates (7,2) what are the coordinates for B?

This is how i set it up

perp bisector ==> neg reciprical
-2 ==> +1/2

y=1/2 x + b
2=1/2 (7) +b
2 = 3.5 + b
b= -1.5
therefore line AB is defined as ===> y = 1/2X - 1.5

set the 2 equations together to get their midpoint
-2x+6 = 1/2x - 1.5
-2.5x = -7.5
2.5x=7.5
x=3

how do we get Y?

perp bisector ==> neg reciprical
-2 ==> +1/2
Is this a rule???

yep, it is a rule.

Cool, thanks:)), did not know this!!!
Re: MGMAT Geometry page 73 #9   [#permalink] 26 Sep 2007, 22:44
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