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Find the equation, in STANDARD FORM, of the line that contains the altitude of Triangle NPQ from point Q to side NP

Points of Triangle NPQ
N (-14, -1)
P (10, -3)
Q (6, 7)
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2 Answers

draw the triangle and label the point of intersection of the altitude to the base NP as point R
the altitude is QR and the base is NP
find the slope of the base NP
N(-14,-1), P(10,-3)
slope=rise/run
slope=(-3-[-1])/(10-[-14])
slope=(-3+1)/(10+14)
slope=-2/24
slope=-1/12 (this is the slope of the base NP)
the slope of the altitude QR, which is perpendicular to the base NP, is the negative reciprocal of the slope of the base
the negative reciprocal of -1/12 is 12
the slope of the altitude QR is 12 and the altitude goes through point (6,7)
y=mx+b
7=12(6)+b
7=72+b
b=7-72
b=-65
y=12x-65 is the equation of the altitude
put this equation in standard form
12x-y=65 is the equation of the altitude
Well, this triangle looks familiar. (:
slopeNP = (-3+1)/(10+14) = -2/24 = -1/12
 
Since the altitude ⊥ NP, their slopes are negative reciprocals of each other. Slope of altitude = 12
 
y-y1=m(x-x1)
y-7 = 12(x-6)
y-7=12x-72
12x - y -65 = 0