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Re M1312
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16 Sep 2014, 00:49
Official Solution:What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4} x  3\) ? A. \(1.4\) B. \(\sqrt{2}\) C. \(1.7\) D. \(\sqrt{3}\) E. \(2.0\) Look at the diagram below: Notice, that the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r= \sqrt{1}=1\). Now, the minimum distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) minus the radius of a circle (which is 1). So we should find the length of the perpendicular, or the height of the right triangle formed by the \(X\) and \(Y\) axis and the line \(y = \frac{3}{4}x3\). The legs of this triangle are: \(leg_1=4\) and \(leg_2=3\). So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\). \(distance=heightradius=2.41=1.4\) Answer: A
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Re: M1312
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06 Oct 2014, 06:37
Bunuel wrote: Official Solution:What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4} x  3\) ? A. \(1.4\) B. \(\sqrt{2}\) C. \(1.7\) D. \(\sqrt{3}\) E. \(2.0\) Look at the diagram below: Notice, that the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r= \sqrt{1}=1\). Now, the minimum distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) minus the radius of a circle (which is 1). So we should find the length of the perpendicular, or the height of the right triangle formed by the \(X\) and \(Y\) axis and the line \(y = \frac{3}{4}x3\). The legs of this triangle are: \(leg_1=4\) and \(leg_2=3\). So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\). \(distance=heightradius=2.41=1.4\) Answer: A I am not comfortable with graphical solution. Is there any way to solve such question without graph.
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Re: M1312
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06 Oct 2014, 07:27
him1985 wrote: Bunuel wrote: Official Solution:What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4} x  3\) ? A. \(1.4\) B. \(\sqrt{2}\) C. \(1.7\) D. \(\sqrt{3}\) E. \(2.0\) Look at the diagram below: Notice, that the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r= \sqrt{1}=1\). Now, the minimum distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) minus the radius of a circle (which is 1). So we should find the length of the perpendicular, or the height of the right triangle formed by the \(X\) and \(Y\) axis and the line \(y = \frac{3}{4}x3\). The legs of this triangle are: \(leg_1=4\) and \(leg_2=3\). So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\). \(distance=heightradius=2.41=1.4\) Answer: A I am not comfortable with graphical solution. Is there any way to solve such question without graph. Check here: whatistheleastpossibledistancebetweenapointonthe85184.html
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Re: M1312
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28 Feb 2015, 20:26
Other solution: Area of right triangle (345): 1/2*3*4 = 6 Since (r+x is height of triangle) => Area = 1/2*(1+x) = 6 => 5x = 7 => x=1.4 Is it right>



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I like it Mrtinhnv.
Another way of finding height of triangle:
Area= 1/2*b*h =1/2*4*3 = 6
therefore: 6=1/2*5*x 12=5*x x=12/5
So 12/51 = 1.4



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Re M1312
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16 Jul 2016, 12:12
I think this is a highquality question.



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Re M1312
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26 Aug 2016, 08:22
I think this is a highquality question and I agree with explanation.



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Re: M1312
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05 Sep 2016, 20:32
After plotting the equations out, I solved it by calculating the Area to find the height (from Origin to the Hypotenuse) Step 1: Draw the circle and the line. You will immediately notice that the triangle is a 345 right triangle. Step 2: Calculate the Area = (3*4)/(2) = 6 Step 3: Calculate the height from the origin to the hypotenuse by using the Area calculated earlier and the hypotenuse as the Base (Base * Height)/(2) = 6 (5 * H)/(2) = 6 (5H) = 12 (H) = (12/5) = 2.4 ***Height = 2.4Step 4: Subtract the radius of 1 from height of 2.4: (2.4  1.0) = 1.4 (answer choice A)
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Re: M1312
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04 Oct 2016, 09:33
Bunuel wrote: Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\).
Hi Bunnel, One doubt in the above piece According to the above theory " The perpendicular... original triangle", if i compare the 'other' smaller triangle with the bigger one, it should still give me the correct equation but I realise, it doesnt in this case: \(\frac{height}{leg_2}=\frac{leg_2}{hypotenuse}\) \(\frac{height}{4}=\frac{4}{5}\) which gives \(height=3.2\) What am i missing?
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Re: M1312
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05 Oct 2016, 02:24
arhumsid wrote: Bunuel wrote: Since the perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\). Hence \(\frac{height}{3}=\frac{4}{5}\), which gives \(height=2.4\).
Hi Bunnel, One doubt in the above piece According to the above theory " The perpendicular... original triangle", if i compare the 'other' smaller triangle with the bigger one, it should still give me the correct equation but I realise, it doesnt in this case: \(\frac{height}{leg_2}=\frac{leg_2}{hypotenuse}\) \(\frac{height}{4}=\frac{4}{5}\) which gives \(height=3.2\) What am i missing? You should consider the ratio of corresponding sides. For similar triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).
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Re: M1312
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11 Oct 2016, 00:03
Hi Bunuel
Any reason why th centre of the circle will necessarily lie on the shortest line drawn between the surface of the circle and the given line?



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Re: M1312
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24 Nov 2016, 18:27
Hi, Bunnel, Can you help me understand why we have drawn the circle with centre as 0 , why not circle with centre other than 0 ?



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Re: M1312
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25 Nov 2016, 00:18



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Re: M1312
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10 Feb 2017, 11:00
Another solution :
Shortest Distance between a point and line is given by:
d = \(\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}\)
Here \(x_{1}\) and \(y_{1}\) is 0,0 (Circle's center ) . Put this value in equation : 3x+4y+12 a=3 , b=4 and c = 3
we will get : 12/5 = 2.4 . Now this is the distance from the center of the circle .So least distance will be from circumference i.e. d  radius => 2.4 1 = 1.4 Ans



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Re: M1312
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15 Feb 2017, 15:56
If we use the property median from right angle is half of the hypotenuse then the height = 5/2 = 2.5. so the answer is 2.51 = 1.5? What am I missing?



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Shiridip wrote: If we use the property median from right angle is half of the hypotenuse then the height = 5/2 = 2.5. so the answer is 2.51 = 1.5? What am I missing? Hi ShiridipThe Shortest distance is the perpendicular distance.Median connects the midpoint of hypotenuse to opposite vertex but is not perpendicular. In case you need a relation between Altitude(h) and sides: h=\(\frac{ab}{c}\), where a and b are the legs measures, and c is the hypotenuse measure. Thanks. +1 Kudos if this post helps .



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Hi Bunuel,
You have mentioned leg1 =4 and leg2 =3. On what basis you have classified leg1 =4 and leg2 =3.
Why not leg1 =3 and leg2 = 4?
Could you please clarify?



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Re: M1312
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02 Jul 2017, 21:57



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Re: M1312
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02 Jul 2017, 22:21
Thanks Bunuel, Now I got it. Anyway it's the multiplication of two lengths, so the order doesn't matter.







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