The line represented by the equation y = 4 - 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?
(A) (-4, 1)
(B) (-2, 2)
(C) (0, 1)
(D) (0, -1)
(E) (2, 0)
OA:D
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pemdas
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perpendicular slope is negative reciprocal of (-2), it's 1/2
the line equation of PR y=(1/2)x+b where b is y-intercept; plug in the point R coordinate to find b, 1=(1/2)*4+b, b=1-2 and b=-1. The line equation of PR is y=(1/2)x-1 and two lines will cross at point (1/2)*x-1=4-2x, (5/2)x=5, x=5*2/5, x=2 and to find y we plug x=2 into the equation y=(1/2)x-1, y=0.
Hence the distance between (4;1) and (2;0) is Sqroot[(4-2)^2+(1-0)^2]=Sqroot(5). We need find another point by knowing the length of PR.
There are two ways to do that both are
1) sketch a graph and find the point below the (2;0);
2) supply the answer choices into (x-2)^2 + (y-0)^2 = 5 and find the suitable choice.
Let's start from the second way,
(A) (-4, 1) --> (-4-2)^2 + 1 = 5 false
(B) (-2, 2) --> (-2-2)^2 + 4 = 5 false
(C) (0, 1) --> (0-2)^2 + 1 = 5 may be
(D) (0, -1) --> (0-2)^2 + 1 = 5 may be
(E) (2, 0) --> (2-2)^2 + 0 = 5 false
to opt among C and D, simply test by using the line equation of PR y=(1/2)x-1
C) (0,1) --> 1=(1/2)*0-1 false
D) (0,-1) --> -1=(1/2)*0-1 true
correct answer is D)
the graphical way is when we just sketch and find the P coordinate on the paper by using our eye measurement, Pythagorean triple and distance calculation (I will not draw it)
the line equation of PR y=(1/2)x+b where b is y-intercept; plug in the point R coordinate to find b, 1=(1/2)*4+b, b=1-2 and b=-1. The line equation of PR is y=(1/2)x-1 and two lines will cross at point (1/2)*x-1=4-2x, (5/2)x=5, x=5*2/5, x=2 and to find y we plug x=2 into the equation y=(1/2)x-1, y=0.
Hence the distance between (4;1) and (2;0) is Sqroot[(4-2)^2+(1-0)^2]=Sqroot(5). We need find another point by knowing the length of PR.
There are two ways to do that both are
1) sketch a graph and find the point below the (2;0);
2) supply the answer choices into (x-2)^2 + (y-0)^2 = 5 and find the suitable choice.
Let's start from the second way,
(A) (-4, 1) --> (-4-2)^2 + 1 = 5 false
(B) (-2, 2) --> (-2-2)^2 + 4 = 5 false
(C) (0, 1) --> (0-2)^2 + 1 = 5 may be
(D) (0, -1) --> (0-2)^2 + 1 = 5 may be
(E) (2, 0) --> (2-2)^2 + 0 = 5 false
to opt among C and D, simply test by using the line equation of PR y=(1/2)x-1
C) (0,1) --> 1=(1/2)*0-1 false
D) (0,-1) --> -1=(1/2)*0-1 true
correct answer is D)
the graphical way is when we just sketch and find the P coordinate on the paper by using our eye measurement, Pythagorean triple and distance calculation (I will not draw it)
sachindia wrote:The line represented by the equation y = 4 - 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?
(A) (-4, 1)
(B) (-2, 2)
(C) (0, 1)
(D) (0, -1)
(E) (2, 0)
OA:D
Last edited by pemdas on Fri Nov 02, 2012 3:25 am, edited 1 time in total.
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GMATGuruNY
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Just draw a picture of what is being described. A perpendicular bisector:sachindia wrote:The line represented by the equation y = 4 - 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?
(A) (-4, 1)
(B) (-2, 2)
(C) (0, 1)
(D) (0, -1)
(E) (2, 0)
OA:D
-- intersects at the midpoint
-- forms a right angle
Here's my rudimentary drawing:
Looking at the drawing above, we can see that P must have a negative y-coordinate.
Only answer choice D works.
The correct answer is D.
Check here for a similar problem:
https://www.beatthegmat.com/grockit-toug ... 74452.html
Last edited by GMATGuruNY on Tue Aug 06, 2013 7:24 pm, edited 1 time in total.
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pemdas
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before just draw the OP mustGMATGuruNY wrote: Just draw a picture of what is being described. A perpendicular bisector:
1) derive the line equation of PR for which one needs to find the slope of perpendicular line
2) use the following rule - perpendicular lines' slopes relate to each other with different signs and reciprocation
3) find the cross-point of two functions (line equations)
4) plot the new coordinates onto x-y coordinate graph
5) apply that the bisector condition
but yea, Mitch, in your eyes it's very simple
the same here after analytic Geo (pre-calculus II), i've done
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There is no need to derive the equation of PR or its point of intersection with y = -2x + 4.pemdas wrote:before just draw the OP mustGMATGuruNY wrote: Just draw a picture of what is being described. A perpendicular bisector:
1) derive the line equation of PR for which one needs to find the slope of perpendicular line
2) use the following rule - perpendicular lines' slopes relate to each other with different signs and reciprocation
3) find the cross-point of two functions (line equations)
4) plot the new coordinates onto x-y coordinate graph
5) apply that the bisector condition
but yea, Mitch, in your eyes it's very simple
the same here after analytic Geo (pre-calculus II), i've done
The purpose of drawing is to AVOID these sorts of calculations.
Here, we need do only the following:
-- sketch y = -2x + 4
-- plot point R at (4,1)
-- draw PR so that it forms a right angle with y = -2x + 4 and is cut in half by y = -2x + 4
A rudimentary drawing is sufficient to see that point P must have a negative y-coordinate, eliminating every answer choice but D.
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Well I really appreciate all the graphical methods but the same may not be applied if the equation changes.
How I did it
when two lines are perpendicular to each other the product of the slopes of the two lines is -1.
m1 * m2 = -1.
Also slope of the bisector (m1) is -2 ( y= mx + c)
so slope (m2) of the line PR should be 1/2 such that product of m1 and m2 is -1.
m2 = change in Y / Change in X. = 1 -y / 4 - x. ( where x, y are the co-ordinates of p)
Plug in the options . And you will get y= -1 and x= 0.
So D.
How I did it
when two lines are perpendicular to each other the product of the slopes of the two lines is -1.
m1 * m2 = -1.
Also slope of the bisector (m1) is -2 ( y= mx + c)
so slope (m2) of the line PR should be 1/2 such that product of m1 and m2 is -1.
m2 = change in Y / Change in X. = 1 -y / 4 - x. ( where x, y are the co-ordinates of p)
Plug in the options . And you will get y= -1 and x= 0.
So D.
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pemdas
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agree, GMAT is not the right place for proofs and axioms 
GMATGuruNY wrote:There is no need to derive the equation of PR or its point of intersection with y = -2x + 4.pemdas wrote:before just draw the OP mustGMATGuruNY wrote: Just draw a picture of what is being described. A perpendicular bisector:
1) derive the line equation of PR for which one needs to find the slope of perpendicular line
2) use the following rule - perpendicular lines' slopes relate to each other with different signs and reciprocation
3) find the cross-point of two functions (line equations)
4) plot the new coordinates onto x-y coordinate graph
5) apply that the bisector condition
but yea, Mitch, in your eyes it's very simple
the same here after analytic Geo (pre-calculus II), i've done
The purpose of drawing is to AVOID these sorts of calculations.
Here, we need do only the following:
-- sketch y = -2x + 4
-- plot point R at (4,1)
-- draw PR so that it forms a right angle with y = -2x + 4 and is cut in half by y = -2x + 4
A rudimentary drawing is sufficient to see that point P must have a negative y-coordinate, eliminating every answer choice but D.
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AbhiJ
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There is no need for graphical or need to find equation of RP.
Let point P be (a,b)
Mid-point of RP will lie on the line y = 4-2x ----1
Slope of RP will = 2, use slope equation for point R and P.
Two equation , two variable.
Let point P be (a,b)
Mid-point of RP will lie on the line y = 4-2x ----1
Slope of RP will = 2, use slope equation for point R and P.
Two equation , two variable.
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pemdas
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Do you mean that a slope of the line segment RP is equal to 2?AbhiJ wrote: Slope of RP will = 2, use slope equation for point R and P.
Two equation , two variable.
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sachindia
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Slope of RP is 1/2 .. as its RP is perpendicular to the other line slope of which is -2
property: m1* m2=-1 ( m1 and m2 being the slopes of 2 perpendicular lines/segments )
property: m1* m2=-1 ( m1 and m2 being the slopes of 2 perpendicular lines/segments )
Regards,
Sach
Sach
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sachindia
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you know that slope of line is -2..pemdas wrote:I did't get this method, anybody please explainAbhiJ wrote:There is no need for graphical or need to find equation of RP.
see my above post.. you will get to know that slope of RP has to be 1/2..
now R is (4,1)
plug in the values from ansswer choices ..
slope=diff in y coordinates/diff in x coordinates..
if you do this,
you will see that for D (0,-1) and R(4,1) ; slope becomes (-1-1)/(0-4) = 1/2 .. So you get to know that D(0,-1 ) is the right answer.. however this method will also lead us to conclude that E is also the rigth answer. .. follow the above and check.
But if you substitute E in the equation of the line, you will find that it lies on the line.. so you can eliminate E.. So D is the ans
Please Thank me if helped
Regards,
Sach
Sach
In the explanation below, I am unable to understand how we are arriving at this step to find the point of intersection of the lines?
The line equation of PR is y=(1/2)x-1 and two lines will cross at point (1/2)*x-1=4-2x
Can someone explain how 1/2)*x-1=4-2x?
Thanks.
The line equation of PR is y=(1/2)x-1 and two lines will cross at point (1/2)*x-1=4-2x
Can someone explain how 1/2)*x-1=4-2x?
Thanks.
pemdas wrote:perpendicular slope is negative reciprocal of (-2), it's 1/2
the line equation of PR y=(1/2)x+b where b is y-intercept; plug in the point R coordinate to find b, 1=(1/2)*4+b, b=1-2 and b=-1. The line equation of PR is y=(1/2)x-1 and two lines will cross at point (1/2)*x-1=4-2x, (5/2)x=5, x=5*2/5, x=2 and to find y we plug x=2 into the equation y=(1/2)x-1, y=0.
Hence the distance between (4;1) and (2;0) is Sqroot[(4-2)^2+(1-0)^2]=Sqroot(5). We need find another point by knowing the length of PR.
There are two ways to do that both are
1) sketch a graph and find the point below the (2;0);
2) supply the answer choices into (x-2)^2 + (y-0)^2 = 5 and find the suitable choice.
Let's start from the second way,
(A) (-4, 1) --> (-4-2)^2 + 1 = 5 false
(B) (-2, 2) --> (-2-2)^2 + 4 = 5 false
(C) (0, 1) --> (0-2)^2 + 1 = 5 may be
(D) (0, -1) --> (0-2)^2 + 1 = 5 may be
(E) (2, 0) --> (2-2)^2 + 0 = 5 false
to opt among C and D, simply test by using the line equation of PR y=(1/2)x-1
C) (0,1) --> 1=(1/2)*0-1 false
D) (0,-1) --> -1=(1/2)*0-1 true
correct answer is D)
the graphical way is when we just sketch and find the P coordinate on the paper by using our eye measurement, Pythagorean triple and distance calculation (I will not draw it)
sachindia wrote:The line represented by the equation y = 4 - 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?
(A) (-4, 1)
(B) (-2, 2)
(C) (0, 1)
(D) (0, -1)
(E) (2, 0)
OA:D
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Jeff@TargetTestPrep
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If the line y = 4 - 2x is the perpendicular bisector of line segment RP, then it's perpendicular to RP at the midpoint of RP. Recall that two lines are perpendicular if their slopes are negative reciprocals of each other. Since the line has a slope of -2, the line segment should have a slope of ½. Thus, let's first determine which answer choice can be point P so that the slope of RP is ½. We'll use the slope formula: m = (y2 - y1)/(x2 - x1):sachindia wrote:The line represented by the equation y = 4 - 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?
(A) (-4, 1)
(B) (-2, 2)
(C) (0, 1)
(D) (0, -1)
(E) (2, 0)
A) (-4, 1)
m = (1 - 1)/(4 - (-4)) = 0
Point P can't be (-4, 1).
B) (-2, 2)
m = (1 - 2)/(4 - (-2)) = -1/6
Point P can't be (-2, 2).
C) (0, 1)
m = (1 - 1)/(4 - 0) = 0
Point P can't be (0, 1).
D) (0, -1)
m = (1 - (-1))/(4 - 0) = 2/4 = 1/2
Point P could be (0, -1).
E) (2, 0)
m = (1 - 0)/(4 - 2) = 1/2
Point P could be (2, 0).
We see that point P could be either (0, -1) or (2, 0), since either one will make RP's slope ½. Next let's determine the midpoint of RP if P is either (0, -1) or (2, 0). We use the midpoint formula: ((x1 + x2)/2 , (y1 + y2)/2)). Recall that R = (4,1).
If P = (0, -1), then the midpoint of RP = ((4 + 0)/2, (1 + (-1))/2) = (2, 0).
If P = (2, 0), then the midpoint of RP = ((4 + 2)/2, (1 + 0)/2) = (3, 1/2).
Recall that the line y = 4 - 2x has to include the midpoint of RP. In other words, the midpoint of RP is a point on the line, and hence its coordinates will satisfy the equation of the line.
If P = (0, -1) and the midpoint of RP is (2, 0), is 0 = 4 - 2(2)? The answer is yes, since 0 = 0. We can see that choice D is the correct choice. However, let's also show that choice E is not the correct choice:
If P = (2, 0) and the midpoint of RP is (3, 1/2), is 1/2 = 4 - 2(3)? The answer is no, since 1/2 ≠-2.
Alternate Solution:
Since the line segment RP is perpendicular to y = 4 - 2x, the line containing RP must have a slope of 1/2, since the slopes of perpendicular lines are negative reciprocals of each other. Then, the line containing the line segment RP must be of the form y = (1/2)x + b, for some real number b. Since this line contains the point R, its equation must be satisfied when we substitute x = 4 and y = 1; therefore 1 = (1/2)(4) + b. Then, 1 = 2 + b, and thus b = -1.
We know the line that contains the line segment RP has an equation y = (1/2)x - 1. Let's find the common point of this line with y = 4 - 2x. We set (1/2)x - 1 = 4 - 2x and solve for x:
(1/2)x -1 = 4 - 2x
(5/2)x = 5
x = 2
So, the x-coordinate of the common point is 2. We can substitute x = 2 in either equation to find the y-coordinate: y = 4 - 2(2) = 0. So, the common point is (2, 0).
Note that this common point is the midpoint of R and P. Therefore, if we let P = (a, b), we must have:
(a + 4)/2 = 2 and (b + 1)/2 = 0. We find that a = 0 and b = -1. Thus, P = (0, -1).
Answer: D
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