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by MBA.Aspirant » Fri Jul 08, 2011 2:42 am
Given line L, and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?
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by Frankenstein » Fri Jul 08, 2011 5:10 am
Hi,
Normal distance of a point(x1,y1) from the line ax+by+c = 0 is |ax1+by1+c|/sqrt(a^2 + b^2)
So, distance of (-1,5) from 3x/4 - y -1/2 =0 is |-3/4-5-1/2|/(5/4) = 5
So, perimeter of rectangle is 2(9+5) = 28
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by amit2k9 » Fri Jul 08, 2011 9:38 pm
equation for the other line is y=3/4x + c
as it passes through (-1,5) hence , 5=3/4*(-1) + c
c=23/4.

hence two lines are y=3/4x-1/2 and y= 3/4+23/4

distance between parallel lines = |23/4-1/2| / [(9/16)+1]^(1/2)

= 25/4 * 4/5 = 5

thus perimeter = 9*2 + 5*2 = 28.
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by amit2k9 » Fri Jul 08, 2011 9:38 pm
equation for the other line is y=3/4x + c
as it passes through (-1,5) hence , 5=3/4*(-1) + c
c=23/4.

hence two lines are y=3/4x-1/2 and y= 3/4+23/4

distance between parallel lines = |23/4-1/2| / [(9/16)+1]^(1/2)

= 25/4 * 4/5 = 5

thus perimeter = 9*2 + 5*2 = 28.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
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