The line represented by the equation y=2x+2 is the perpendicular bisector of of line segment MN. if the coordinates of point N are (1,-1), wat r the cordinates of point M?
1. (-2, 1)
2. (-3, 1)
3. (-2.5, 0.75)
4. (-2, -1)
5. (-2, 0.75)
The line represented by the equation y=2x+2 is the perpendicular bisector of of line segment MN. if the coordinates of point N are (1,-1), wat r the cordinates of point M?
1. (-2, 1)
2. (-3, 1)
3. (-2.5, 0.75)
4. (-2, -1)
5. (-2, 0.75)
Slope of line MN should be -1/2 since the slope of perpendicular bisec is 2.
Option B & option C statisfy the condition.Midpoint of line MN must pass thru the perpendicular bisec.Only option B satisfies the condition.
I will go with option B
-Deepak.
Slope of line MN should be -1/2 since the slope of perpendicular bisec is 2.
Option B & option C statisfy the condition.Midpoint of line MN must pass thru the perpendicular bisec.Only option B satisfies the condition.
I will go with option B
-Deepak.
hi deepak.. could u elaborate the bold text above?? how can option C satisfy? n how only option B satisfies the condition?
dare2 Sayshi deepak.. could u elaborate the bold text above?? how can option C satisfy? n how only option B satisfies the condition?
Can't give a better explanation for this than Ankit.
Though the question is not directed at me, here is how to solve this:
Given Line is y=2x+2...........(1)
Let the perpendicular bisector have the equation y=mx+c.....(2)
Product of Slopes of perpendicular lines =-1
Thus,
m.2=-1 => m=-1/2..................(3)
Point N(1,-1) lies on this line and thus should satisfy equation (2).
Thus,
-1=-1/2+c => c= -1/2......(4)
From (2),(3) AND (4)
2y=-x-1...........................(5)
Point M will always lie on this line and thus should satisfy (5). Only option B and C do so.
Now, the midpoint of MN will be intersection of lines (1) and (5). Solving, we get point P (-1,0).
Let have co-ordinates (x,y).
Length of PM=Length PN= sqrt(5). .................................(6)
Substitute values from B and C in (6) to get your answer as B.
nice explanation ankit..
but actually i was looking 4 a shorter, smarter solution.. i tried by plotting the line n points in x-y plane.. but my method was riskier coz options were very close and also coz my drawing was not so accurate.. so i was looking if someone could come up wid some direct solutions..
The shortest way of doing this is taking the answers into finding the solution
y=2x+2 --------------(1) and since it was given it is the perpendiculare bisector of the line the distance between M to the line(1) and N to the line is same which is the perpendicular distance from the point to a line
d=(2+1+2)/(sqrt(5)) which is sqrt(5)
substitute the given options in the eqn and which one gives sqrt(5) gives the soln.
In this case, the time taken is very less as we have to do only addition in numerator because denmoinator will be same for any of the points.
regards,
Sandeep Reddy.
The shortest way of doing this is taking the answers into finding the solution
y=2x+2 --------------(1) and since it was given it is the perpendiculare bisector of the line the distance between M to the line(1) and N to the line is same which is the perpendicular distance from the point to a line
d=(2+1+2)/(sqrt(5)) which is sqrt(5)
substitute the given options in the eqn and which one gives sqrt(5) gives the soln.
In this case, the time taken is very less as we have to do only addition in numerator because denmoinator will be same for any of the points.
regards,
Sandeep Reddy.
The shortest way of doing this is taking the answers into finding the solution
y=2x+2 --------------(1) and since it was given it is the perpendiculare bisector of the line the distance between M to the line(1) and N to the line is same which is the perpendicular distance from the point to a line
d=(2+1+2)/(sqrt(5)) which is sqrt(5)
substitute the given options in the eqn and which one gives sqrt(5) gives the soln.
Ans Is B
In this case, the time taken is very less as we have to do only addition in numerator because denmoinator will be same for any of the points.
regards,
Sandeep Reddy.
The line represented by the equation y=2x+2 is the perpendicular bisector of of line segment MN. if the coordinates of point N are (1,-1), what r the coordinates of point M?
1. (-2, 1)
2. (-3, 1)
3. (-2.5, 0.75)
4. (-2, -1)
5. (-2, 0.75)
There are two conditions:
- Line parallel to y=2x+2 implies slope = -1/2
to apply this is easy. The slope is (-1-Y)/(1-X)
Eliminates : 1, 4, 5.
Left with 2 & 3.
- M equidistant from the point of intersection.
Equation of MN => y+1=(-1/2)(x-1)
=> 2y = -x-1.
Solving these eqtns:
2y = 4x + 4 ...(a)
2y = -x -1 ...(b)
(a) - (b) => 5x+5=0 => x=-1.
Y= -2+2 = 0. Midpoint = (-1,0). Call it P. PN = ((-1-1)^2 + (0-1)^2)^1/2 = sqrt(5).
i.e. MP=sqrt(5).
Now using the option 2 : MP = sqrt(sq(-3-1)+sq(1-0)) = sqrt(5). Correct.
using choice (3) : MP = sqrt(sq(-2.5-1)+sq(.75-0))
hence choice (2) is correct.
The line represented by the equation y=2x+2 is the perpendicular bisector of of line segment MN. if the coordinates of point N are (1,-1), wat r the cordinates of point M?
1. (-2, 1)
2. (-3, 1)
3. (-2.5, 0.75)
4. (-2, -1)
5. (-2, 0.75)
Good one. Looks typically 'NON-GMAT'ical ;)
Option 2 😃
Please help me with this Gmat Prep question:
For every positive integer n,h(n) is the product of even numbers from 2 to n,inclusive.If P is the smallest prime factor of h(100)+1,then P is:
1)between 2 and 10
2)between 10 and 20
3)between 20 and 30
4)between 30 and 40
5)greater than 40
Please help me with this Gmat Prep question:
For every positive integer n,h(n) is the product of even numbers from 2 to n,inclusive.If P is the smallest prime factor of h(100)+1,then P is:
1)between 2 and 10
2)between 10 and 20
3)between 20 and 30
4)between 30 and 40
5)greater than 40
use Mission's theorem.. :)
check previous pages..
if a number is selected from 10000 to 99999 inclusive, wat is the probability of getting a number that contains four digits of '5'?
(not putting the options intentionally, else u can pick the rite answer within 10 seconds)
if a number is selected from 10000 to 99999 inclusive, wat is the probability of getting a number that contains four digits of '5'?
(not putting the options intentionally, else u can pick the rite answer within 10 seconds)
IMO 1/2000 whats the OA?
cannot delete the post
vpitc SaysIMO 1/2000 whats the OA?
plz post ur approach wid explanation..
if a number is selected from 10000 to 99999 inclusive, wat is the probability of getting a number that contains four digits of '5'?
(not putting the options intentionally, else u can pick the rite answer within 10 seconds)
Total sample space = 90,000
consider 5 digits number abcde
case1: a={1,9} - {5} so bcde=5555 :total number of way =8
case2: fixed 5 at 5th digit and then 4th n so on
its like
9 ,9 , 9,and 10
total number of ways = 8+9+9+9+10 =45
P = 45/90,000 = 1/2,000
Total sample space = 90,000
consider 5 digits number abcde
case1: a={1,9} - {5} so bcde=5555 :total number of way =8
case2: fixed 5 at 5th digit and then 4th n so on
its like
9 ,9 , 9,and 10 (10 cases wont b possible coz all the digits will become 5.. in question we r asked 2 find the number of four 5's not five 5's)
total number of ways = 8+9+9+9+10 =45
P = 45/90,000 = 1/2,000
dude.. u juz missed by a whisker.. another method will be
no. of ways of choosing 4-digits 4m 5 digits(to fill wid 5's) = 5C4 = 5
the fifth digits can be filled in 9 ways... (0,1 to 9 but widout 5)
so total ways = 5x9=45
but in these 45 ways there was 1 case in which 0 was the first digit.. so we need 2 subtract 1 from 45..
so total eligible cases = 44
probability =44/90000
Try this,
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
a 28
b 32
c 48
d 60
e 120
Try this,
A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
a 28
b 32
c 48
d 60
e 120
when father drives.. the 4 seats can be arranged(seated) in 4! ways = 24
OR when mother drives.. the 4 seats...........................................= 24
total ways = 48
but in the above 48 cases, we counted the cases when the 2 daughters were sitting together.. so we need 2 subtract those case..
there were 2 cases in when father drove n 2 cases when mother drove..
_ _ _
__ _
also the two daughters can swap positions between themselves..
unwanted cases when father drove = 2 x 2 = 4
AND unwanted cases when mother drove = 2 x 2 = 4
total unwanted cases = 4 x 4 = 16
possible seating arrangements = 48-16 =32
option B..