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Kaplan: Smallest Distance on a co-ordinate plane

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by shubhamkumar » Mon Apr 16, 2012 8:41 pm
Solution:
The Line segment y=4/3x-100 will intersect the x axis on (75,0) and Y axis on (0,-100).SO this line segment will lie in the 4th Quadrant(though this has nothing to do with the soln)
The shortest distance will be the perpendicular drawn on the line segment from the origin.
Since the slope of perpendiculars are negative reciprocals,the slope of this perpendicular will be -3/4.
And hence the equation of this perpendicular will be y =(-3/4)x.
Next find the point of intersection of this perpendicular to the line segment in question.
y=4/3x-100
or -3/4x=4/3-100
or 25/12x=100
or x=48
so y=-36
(48,-36) is the closest point on the line segment to the origin.
To find the distance you can use the formula distance=((x1-x2)^2+(y1-y2)^2)^1/2
Or use the quicker approach mentioned in Kaplan

If we draw a pythagrous triangle with the x axis,line segment y=4/3-100 and y=-3/4x.The length of the two sides of the triangle will be in the ratio 4:3=12(4):12(3),which are the 2 sides of the pythagorous triplet 3:4:5.Hence the distance of this similar triangle will be 5 and the distance in the original triangle will be 5*12=60.[spoiler](C)[/spoiler]
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by aneesh.kg » Mon Apr 16, 2012 8:42 pm
The smallest distance between a point and a line is the length of the perpendicular line segment drawn from the point onto the line.

So, this is what we will do:
1. Draw a perpendicular from the origin onto the given line.
2. Find the equation of the perpendicular line.
Slope = -1 /(4/3) = -3/4
Since it passes through the origin, the equation is
y = (-3/4)x
3. Find the point of intersection of the two lines
By solving the two equations, we get the point of intersection = (48,-36)
4. Find the distance of this point from the origin (which must be equal to the required smallest distance)
d = (48^2 + 36^2)^0.5 = 12.(4^2 + 3^2)^0.5 = 60

[C] is the answer
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by killer1387 » Mon Apr 16, 2012 8:50 pm
shubhamkumar wrote:The equation of line n is y = 4/3x - 100. What is the smallest possible distance in the xy-plane from the point with coordinates (0, 0) to any point on line n?

A.48
B.50
C.60
D.75
E.100

OA: C
3y = 4x - 300

4x-3y=300

shortest distance = 300/sq rt.(16+9)=300/5= 60

hence C
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by jaijune » Mon Apr 16, 2012 9:07 pm
If ax+by+c =0 then shortest distance from origin =|c|/(a^2+b^2)^1/2

The equation of line n is y = 4/3x - 100
4x-3y-300=0

a=4, b=-3, c=-300
If we substitute these values will get the answer as 60.
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