Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 09 Oct 2015, 21:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Given the two lines y = 2x + 5 and y = 2x - 10, what is the

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Manager
Joined: 19 Oct 2011
Posts: 135
Location: India
Followers: 2

Kudos [?]: 250 [5] , given: 33

Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  02 Sep 2012, 19:05
5
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

61% (03:34) correct 39% (02:05) wrong based on 132 sessions
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

A. $$\frac{45}{4}*\pi$$

B. $$2\sqrt{45}\pi$$

C. $$27\pi$$

D. $$27\sqrt{2}\pi$$

E. $$45\pi$$
[Reveal] Spoiler: OA

_________________

Encourage me by pressing the KUDOS if you find my post to be helpful.

Help me win "The One Thing You Wish You Knew - GMAT Club Contest"
the-one-thing-you-wish-you-knew-gmat-club-contest-140358.html#p1130989

Last edited by Bunuel on 22 Aug 2014, 20:31, edited 1 time in total.
Edited the question.
Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 81

Kudos [?]: 670 [4] , given: 43

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  29 Sep 2012, 13:23
4
KUDOS
dvinoth86 wrote:
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

[Reveal] Spoiler:
(45/4)pi

Even if you don't know the formula for the distance between two parallel lines, you can figure it out.

The two lines are parallel, having the same slope 2.
A circle tangent to both lines will have the diameter equal to the distance between the two lines. To find this distance,
take the perpendicular from the origin to each of the lines.
The origin and the x-intercept and the y-intercept of the line $$y = 2x - 10$$ form a right triangle with legs 5 and 10 and hypotenuse $$5\sqrt{5}$$.
Therefore, the height corresponding to the hypotenuse is $$10\cdot{5}/5\sqrt{5}=2\sqrt{5}$$ (use the formula height = leg*leg/hypotenuse).
The distance between the origin and the line $$y = 2x + 5$$ is half of the distance between the origin and the line $$y = 2x - 10$$, because the x-intercept and the y-intercept of the line $$y = 2x + 5$$ are -5/2 and 5, which with the origin, form a right triangle similar to the right triangle discussed above.

So, the diameter of the circle is $$3\sqrt{5}$$ and the area of the circle is $$\pi\cdot9\cdot{5}/4=(45/4)\pi$$.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Status: Tutor
Joined: 05 Apr 2011
Posts: 537
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 570 Q49 V19
GMAT 2: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Followers: 74

Kudos [?]: 395 [3] , given: 47

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  02 Sep 2012, 19:17
3
KUDOS
1
This post was
BOOKMARKED
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

Diameter of the circle will be equal to the distance between these two parallel tangents.

Distance between parallel lines is given by the formula
d = |C1–C2|/√(A^2+B^2)
(refer: http://www.askiitians.com/iit_jee-Strai ... llel_lines)

Here C1 = 5, C2=-10
A = 2
B = -1

So d = |5–(-10)|/√(2^2+(-1)^2)
=> d = 3√5

So area = (pie d^2)/4
= pie * (3√5(^2) )/ 4
= 45 pie /4

Hope it helps!
_________________
Intern
Joined: 12 Jun 2012
Posts: 42
Followers: 1

Kudos [?]: 23 [2] , given: 28

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  09 Oct 2012, 07:09
2
KUDOS
My answer by using first principles and not the A and B formula

Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

The first thing to use is the perpendicular bi-sector between the two lines. If y=2x is the gradient of the lines, then y= -1/2x is the gradient of the bisector.

Step 1, solve the two simultaneous equations in order to find the vectors of the line you seek.

y=2x+5
y= -1/2x
x = -2
Subbing -2 into y=2x+5=1
So co-ord 1 is (-2,1)

Next
y=2x-10
y=-1/2x
x=4
Subbing 4 into y=2x-10=-2
Co-ord to is (4,-2)

Now you have a pythagorus equation
Triangle base = 6
Triangle height = 3

Hypotenuse (the diameter of the circle) ^2 = 36 + 9
Hyp= diameter = sqrt(45)
Area = Pi x r^2
=45/4 x pi
_________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 81

Kudos [?]: 670 [1] , given: 43

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  15 Oct 2012, 10:03
1
KUDOS
IanSolo wrote:
nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: $$d = \frac{15}{\sqrt{5}}$$
Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different.
I'm not able to find the error.

$$\frac{15}{\sqrt{5}}=\frac{15\sqrt{5}}{\sqrt{5}\sqrt{5}}=\frac{15\sqrt{5}}{5}=3\sqrt{5}$$ which is the same as $$\sqrt{9\cdot{5}}=\sqrt{45}$$.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Math Expert
Joined: 02 Sep 2009
Posts: 29802
Followers: 4905

Kudos [?]: 53661 [1] , given: 8167

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  23 Aug 2014, 11:10
1
KUDOS
Expert's post
desaichinmay22 wrote:
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

Point (0,5) satisfies equation y=2x+5
Point (5,0) satisfies equation y=2x-10
Take (0,0) to form a right triangle with points (0,5) and (5,0)
Height of triangle= distance between (0,0) and (0,5) = 5
Base of triangle=distance between (0,0) and (5,0) = 5
Therefore hypotenuse =5\sqrt{2}
That is the diameter of the circle.
Area of circle= pi*r^2
=pi*(5\sqrt{2}/2)^2
=pi*50/4

Can anyone explain what am I missing in this approach.

The point is that a circle tangent to both those lines must have the diameter equal to the distance between the two lines. The distance between two parallel lines is the length of perpendicular from one to another, while line segment joining (5,0) and (0,5) is NOT perpendicular to the lines:
Attachment:

Untitled.png [ 12.54 KiB | Viewed 1566 times ]
So, this line segment cannot be the diameter of the circle.

Hope it's clear.
_________________
Intern
Joined: 09 Sep 2012
Posts: 33
Location: United States
Followers: 2

Kudos [?]: 10 [0], given: 33

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  29 Sep 2012, 12:23
nktdotgupta wrote:
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

Diameter of the circle will be equal to the distance between these two parallel tangents.

Distance between parallel lines is given by the formula
d = |C1–C2|/√(A^2+B^2)
(refer: http://www.askiitians.com/iit_jee-Strai ... llel_lines)

Here C1 = 5, C2=-10
A = 2
B = -1

So d = |5–(-10)|/√(2^2+(-1)^2)
=> d = 3√5

So area = (pie d^2)/4
= pie * (3√5(^2) )/ 4
= 45 pie /4

Hope it helps!

Hi all,

Pls explain the question above !!!
What is a and b in he above formula ?
Intern
Joined: 09 Oct 2012
Posts: 39
Followers: 0

Kudos [?]: 4 [0], given: 14

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  15 Oct 2012, 09:44
nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: $$d = \frac{15}{\sqrt{5}}$$
Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different.
I'm not able to find the error.
Status: Tutor
Joined: 05 Apr 2011
Posts: 537
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 570 Q49 V19
GMAT 2: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Followers: 74

Kudos [?]: 395 [0], given: 47

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  16 Oct 2012, 05:58
IanSolo wrote:
nktdotgupta I have tried to solve with your way, but when I try to calculate d I receive another result: $$d = \frac{15}{\sqrt{5}}$$
Then I divide d with 2 for find r and eventually I elevate r^2 and moltiplicate with pie, but the result is different.
I'm not able to find the error.

d = 15/sqrt5 = 3sqrt5

area = (pie * d^2)/4
= (pie * (3sqrt5)^2)/4
= (pie * 45/4)

hope it helps!
_________________
Current Student
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1099
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Followers: 34

Kudos [?]: 388 [0], given: 69

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  11 Nov 2012, 14:42
nktdotgupta wrote:
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

Diameter of the circle will be equal to the distance between these two parallel tangents.

Distance between parallel lines is given by the formula
d = |C1–C2|/√(A^2+B^2)
(refer: http://www.askiitians.com/iit_jee-Strai ... llel_lines)

Here C1 = 5, C2=-10
A = 2
B = -1

So d = |5–(-10)|/√(2^2+(-1)^2)
=> d = 3√5

So area = (pie d^2)/4
= pie * (3√5(^2) )/ 4
= 45 pie /4

Hope it helps!

Hi

a very fundamental problem with the explanation.

Had the two eqns of the form

y-2x-5 = 0
and y - 2x +10 =0

which is same to the above eqn, should the eqn be converted to y = mx + c form for getting c1 and c2

i.e the eqn Ax + By + C = 0 needs to be converted to form y = mx+ c or not.
Intern
Joined: 03 Jun 2013
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  31 Jul 2013, 03:13
@bunuel can u pls help n explain . i cant understand either
Intern
Joined: 17 May 2013
Posts: 49
GMAT Date: 10-23-2013
Followers: 0

Kudos [?]: 6 [0], given: 8

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  02 Aug 2013, 06:30
to clarify what nkdotgupta wrote

write the equation in the form of ax+by+c

Eq 1. y = 2x + 5 can be written as 2x-y+5 = 0 which makes a = 2 b = -1 and c1 = 5

Eq 2. y = 2x - 10 can be written as 2x-y-10 = 0 which makes a = 2 b = -1 and c2 = -10

hope this helps..
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 6783
Followers: 367

Kudos [?]: 83 [0], given: 0

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  22 Aug 2014, 14:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 21 Sep 2012
Posts: 153
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Followers: 1

Kudos [?]: 106 [0], given: 31

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  23 Aug 2014, 10:20
Point (0,5) satisfies equation y=2x+5
Point (5,0) satisfies equation y=2x-10
Take (0,0) to form a right triangle with points (0,5) and (5,0)
Height of triangle= distance between (0,0) and (0,5) = 5
Base of triangle=distance between (0,0) and (5,0) = 5
Therefore hypotenuse =5\sqrt{2}
That is the diameter of the circle.
Area of circle= pi*r^2
=pi*(5\sqrt{2}/2)^2
=pi*50/4

Can anyone explain what am I missing in this approach.
Manager
Joined: 21 Sep 2012
Posts: 153
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Followers: 1

Kudos [?]: 106 [0], given: 31

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  23 Aug 2014, 12:00
Bunuel wrote:
desaichinmay22 wrote:
Given the two lines y = 2x + 5 and y = 2x - 10, what is the area of the largest circle that can be inscribed such that it is tangent to both lines

Point (0,5) satisfies equation y=2x+5
Point (5,0) satisfies equation y=2x-10
Take (0,0) to form a right triangle with points (0,5) and (5,0)
Height of triangle= distance between (0,0) and (0,5) = 5
Base of triangle=distance between (0,0) and (5,0) = 5
Therefore hypotenuse =5\sqrt{2}
That is the diameter of the circle.
Area of circle= pi*r^2
=pi*(5\sqrt{2}/2)^2
=pi*50/4

Can anyone explain what am I missing in this approach.

The point is that a circle tangent to both those lines must have the diameter equal to the distance between the two lines. The distance between two parallel lines is the length of perpendicular from one to another, while line segment joining (5,0) and (0,5) is NOT perpendicular to the lines:
Attachment:
Untitled.png
So, this line segment cannot be the diameter of the circle.

Hope it's clear.

Hi Bunuel,

I got your point. Can you please suggest how to derive the length of perpendicular using graphical approach.
I am not able to get it.
Senior Manager
Joined: 23 Jan 2013
Posts: 423
Schools: Cambridge'16
Followers: 2

Kudos [?]: 33 [0], given: 34

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  24 Aug 2014, 03:50
I used the following formula to find distance between lines:
|ax+by+c|/sqrt(a^2+b^2), got 15/sqrt5,
halved and got 15/2*(sqrt5),
squared it to get 45/4pi

But it took more than 5 min. to solve
Manager
Joined: 04 Oct 2013
Posts: 162
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Followers: 3

Kudos [?]: 77 [0], given: 54

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  24 Aug 2014, 06:21
1
This post was
BOOKMARKED
Graphical approach:

Straight lines with equation f(x) = 2x + 5 and g(x) = 2x - 10 are parallel; slope of each line equal to 2.

Co-ordinate of y-intercepts of lines f(x) and g(x) are (0,5) and (0,-10)

Equation of the line perpendicular to f(x) and g(x) and passing through y-intercept of g(x):h(x) = -(1/2)x-10

Point of intersection between h(x) and f(x) : (-6,-7)
[ Point of intersection satisfies both line equations f(x) and h(x). Thus, $$2x+5=\frac{-1}{2}x-10$$ or$$x =-6$$ and $$y=-7$$]

Distance between parallel lines = Distance between (-6,-7) and (0,-10) = $$\sqrt{( -10+7)^2+(0+6)^2}$$$$=3\sqrt{5}$$

Largest circle that can be inscribed between lines f(x) and g(x) must have the diameter equal to the distance between these two lines = $$3\sqrt{5}$$

Area of the circle = $$\pi*(3\sqrt{5}/2)^2= \frac{45}{4}*\pi$$

Attachments

Distance-Between-Parallel-Lines.png [ 24.44 KiB | Viewed 1383 times ]

Intern
Joined: 27 Aug 2014
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 17

Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]  28 Aug 2014, 05:59
Temurkhon wrote:
I used the following formula to find distance between lines:
|ax+by+c|/sqrt(a^2+b^2), got 15/sqrt5,
halved and got 15/2*(sqrt5),
squared it to get 45/4pi

But it took more than 5 min. to solve

yes, I did the same thing. 5 min. Too bad
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the   [#permalink] 28 Aug 2014, 05:59
Similar topics Replies Last post
Similar
Topics:
2 The line represented by the equation y = - 2 x + 6 is 3 20 Sep 2013, 15:17
If 2x = 3y = 10, then 12xy = ? 1 17 Sep 2013, 16:52
19 If x+y=2 and x^2 - xy - 10 - 2y^2 = 0, what does x-2y =? 10 28 Feb 2012, 20:12
22 The line represented by the equation y = 4 – 2x is the 28 18 Feb 2012, 17:55
1 If 2x + 5y =8 and 3x = 2y, what is the value of 2x + y? 2 28 Mar 2010, 08:32
Display posts from previous: Sort by

# Given the two lines y = 2x + 5 and y = 2x - 10, what is the

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.