SOLUTION: What is the equation "in slope-intercept form" of the line perpindicular to the segment between (1,3) and (5,5) and passing through (3,4)

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Question 478306: What is the equation "in slope-intercept form" of the line perpindicular to the segment between (1,3) and (5,5) and passing through (3,4)
Answer by lwsshak3(11628) About Me  (Show Source):
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What is the equation "in slope-intercept form" of the line perpindicular to the segment between (1,3) and (5,5) and passing through (3,4)
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Standard form for a straight line: y=mx+b, m=slope, b=y-intercept
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Slope of line segment between (1,3) and (5,5)=∆y/∆x=(5-3)/5-1=2/4=1/2
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Slope of line perpendicular to line segment=-2 (negative reciprocal)
Equation of line: y=-2x+b
solving for b using given point line passes thru, (3,4)
4=-2*3+b
4=-6+b
b=10
Equation: y=-2x+10