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07 Nov 2019, 02:19
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Difficulty:
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What is the smallest possible distance between origin and any point on the line \(y=\frac{1}{2}x+50\)?
A. 25
B. \(20\sqrt{5}\)
C. 50
D. \(50\sqrt{2}\)
E. 100
Are You Up For the Challenge: 700 Level Questions
A. 25
B. \(20\sqrt{5}\)
C. 50
D. \(50\sqrt{2}\)
E. 100
Are You Up For the Challenge: 700 Level Questions
09 Nov 2019, 06:04
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Bunuel
There can be many ways to solve it..
(I) Direct formula....
If the line is ax+by+c=0, then the shortest/perpendicular distance of point (X,Y) is \(\frac{aX+bY+c}{\sqrt{a^2+b^2)}}\)
Now \(y=\frac{1}{2}x+50......2y=x+100......x-2y+100=0\), so a=1, b=-2, c=100 and (X,Y)=(0,0)
\(\frac{aX+bY+c}{\sqrt{a^2+b^2}}\)=\(\frac{0+0+100}{\sqrt{1^2+(-2)^2}}=\)\(\frac{100}{\sqrt{5}}=20\sqrt{5}\)
(II) Slope
Now slope in \(y=\frac{1}{2}x+50\) is 1/2, so the slope of the perpendicular will be -2 as the product of slopes of perpendicular lines is -1.
The equation of the line passing through (0,0) and of slope -2 is...\(\frac{y-y_1}{x-x_1}=m.......\frac{y-0}{x-0}=-2.....y=-2x\)
Now where will the two lines meet or the perpendicular from (0,0) meet the line \(y=\frac{1}{2}x+50\).....\(y=-2x=\frac{1}{2}x+50...x=-20\)
and y=-2x=-2*20=-40..
Distance between (0,0) and (-20,-40)=\(\sqrt{(-20-0)^2+(-40-0)^2}=\sqrt{20^2+40^2}=20\sqrt{1^2+2^2}=20\sqrt{5}\)
(II) Graph..
Draw a free hand graph as shown..
The area of \(\triangle{ABO}=\frac{1}{2}*50*100=2500\)...(i)
The area of \(\triangle{ABO}=\frac{1}{2}*AB*OC\), when AB is taken as the BASE..
AB=\(\sqrt{100^2+50^2}=50\sqrt{2^2+1^2}=50\sqrt{5}\), so Area = \(\frac{1}{2}*50\sqrt{5}*OC=25\sqrt{5}*OC\)...(ii)
From (i) and (ii), we get \(25\sqrt{5}*OC=2500....OC=\frac{2500}{25\sqrt{5}}=20\sqrt{5}\)
B
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General Discussion
07 Nov 2019, 09:12
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Equation of a line is y=mx+c
If a line passes from the origin then the equation of line is of the form y=mx
Here the given line is y= (1/2)x + 50, which gives us slope=1/2
So slope of line perpendicular to it is -2 as the product of the slopes =-1 in case of perpendicular lines.
So equation of line from origin is y=-2x
Let's find the point of intersection by solving the two simultaneous equations.
-2x = (1/2)x +50
-2.5x = 50
x = -20
y=-2x=40
So the distance from origin, (0,0) to (-20,40) is
Sqrt( (-20)^2 + (40)^2 )
= 20sqrt(1+4)
=20 sqrt(5)
Posted from my mobile device
If a line passes from the origin then the equation of line is of the form y=mx
Here the given line is y= (1/2)x + 50, which gives us slope=1/2
So slope of line perpendicular to it is -2 as the product of the slopes =-1 in case of perpendicular lines.
So equation of line from origin is y=-2x
Let's find the point of intersection by solving the two simultaneous equations.
-2x = (1/2)x +50
-2.5x = 50
x = -20
y=-2x=40
So the distance from origin, (0,0) to (-20,40) is
Sqrt( (-20)^2 + (40)^2 )
= 20sqrt(1+4)
=20 sqrt(5)
Posted from my mobile device
