How do I find this?
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The shortest distance will be from a point on the given line that is perpendicular to it and contain the point (0,0)
a line perpendicular to y=3x-5 will have a slope that is the negative reciprocal of the given line
slope of given line is 3
slope of perpendicular line is (-1/3) so the equation of the line is
y=(-1/3)x+b with point (0,0) on this line
so 0=(-1/3)x+b b=0
equation of this line is y=(-1/3)x
solving the two equations will given you the point on the given line where they cross.
y=3x-5 and y=(-1/3)x
substituting you have
(-1/3)x=3x-5:
-(10/3)x=-5
x=15/10=3/2
In the original equation the value of y when x=(3/2) is
y=3(3/2)-5 = 9/2 -10/2
y=(-1/2)
the point on the original line that is closest to the original is
Point[(3/2),(-1/2)]
The distance from the original to this point is
√[(x1-x2)^(2) + (y1-y2)^(2)]
√ [(3/2)-0]^2 + [(-1/2)-0]^2]=
√[ (9/4)+(1/4)]=
√ (10/4) = √(5/2) = √(10) / 2
dnadan1
a line perpendicular to y=3x-5 will have a slope that is the negative reciprocal of the given line
slope of given line is 3
slope of perpendicular line is (-1/3) so the equation of the line is
y=(-1/3)x+b with point (0,0) on this line
so 0=(-1/3)x+b b=0
equation of this line is y=(-1/3)x
solving the two equations will given you the point on the given line where they cross.
y=3x-5 and y=(-1/3)x
substituting you have
(-1/3)x=3x-5:
-(10/3)x=-5
x=15/10=3/2
In the original equation the value of y when x=(3/2) is
y=3(3/2)-5 = 9/2 -10/2
y=(-1/2)
the point on the original line that is closest to the original is
Point[(3/2),(-1/2)]
The distance from the original to this point is
√[(x1-x2)^(2) + (y1-y2)^(2)]
√ [(3/2)-0]^2 + [(-1/2)-0]^2]=
√[ (9/4)+(1/4)]=
√ (10/4) = √(5/2) = √(10) / 2
dnadan1
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3x-y-5 = 0
s = |-5|/sqrt(10) = sqrt(10)/2
s = |-5|/sqrt(10) = sqrt(10)/2