Question 175131: I need help with the following
- Find the shortest distance from the point (3,1) to the line 3x+y=-2 to the nearest tenth.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I need help with the following
- Find the shortest distance from the point (3,1) to the line 3x+y=-2 to the nearest tenth.
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Draw the picture.
Find the equation of the line perpendicular to the given line and passing
thru (3,1)
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slope of the given line: -3
slope of the perpendicular line: 1/3
intercept for this line: 1 = (1/3)3 + b
b = 0
Equation of the perpendicular line: y = (1/3)x
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intercept of the given line and the perpendicular line:
y = (1/3)x
y = -3x-2
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Substitute to solve for "x": (1/3)x = -3x-2 ; (8/3)x=-2;x =-3/4
Since y = (1/3)x y = (1/3)(-3/4) = -1/4
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So the intercept point is (-3/4,1/4)
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Find the distance from (-3/4,-1/4) t0 (3,1)
distance = sqrt[(1--1/4)^2 + (3--3/4)^2] = sqrt[(25/16) + (225/16)]
= sqrt[250/16] = (5/4)sqrt(10)
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Cheers,
Stan H.
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