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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.
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]
09 Oct 2012, 07:09
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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) Radius = Sqrt(45)/2 Area = Pi x r^2 =45/4 x pi _________________
If you find my post helpful, please GIVE ME SOME KUDOS!
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.
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]
15 Oct 2012, 10:03
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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.
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.
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!
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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.
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]
23 Aug 2014, 11:10
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Expert's post
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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 2164 times ]
So, this line segment cannot be the diameter of the circle.
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.
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
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 = \(\pi*(3\sqrt{5}/2)^2= \frac{45}{4}*\pi\)
Answer: (A)
Attachments
Distance-Between-Parallel-Lines.png [ 24.44 KiB | Viewed 1979 times ]
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
Re: Given the two lines y = 2x + 5 and y = 2x - 10, what is the [#permalink]
05 Dec 2015, 11:03
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. _________________
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