Question 199381
 Determine the shortest distance from the point H(5,2) tot he line through points J(-6,4) and K(-2,-4) 
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Method:  Find the slope of the line JK.  The shortest distance from a point to a line is along a line perpendicular, to JK in this case.  The slope of lines perpendicular is the negative inverse of the slope of JK.
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Slope, m, of JK = diffy/diffx = (-4-4)/(-2 - (-6)) = -8/4
m of JK = -2
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The slope of ALL lines perpendicular to JK have a slope of +1/2
To find the equation of the line thru H with a slope of +1/2, use
y-y1 = m*(x-x1)
y-2 = (1/2)*(x-5)
2y-4 = x-5
x-2y = 1  (eqn of line thru H)
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Find the eqn of JK:
Use the same method with either point, J or K
y+4 = -2*(x+2)
y = -2x-8
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Find where the lines intersect by solving the 2 eqns.
x-2y = 1  (eqn of line thru H)
y = -2x-8
Sub for y into eqn thru H
x - 2(-2x-8) = 1
x+4x + 16 = 1
5x = -15
x = -3
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y = -2
--> Intersection at (-3,-2)
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Find distance from H to intersection:
d^2 = (5+3)^2 + (2+2)^2 = 64+16
distance = sqrt(80)
{{{d = 4sqrt(5)}}}