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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.
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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

1

This post received 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) Radius = Sqrt(45)/2 Area = Pi x r^2 =45/4 x pi
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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

1

This post received 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.

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.