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

Question

Points of Triangle NPQ

N (-14, -1)

P (10, -3)

Q (6, 7)

Answer 1

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

Answer 2

Well, this triangle looks familiar. (:

slope_NP = (-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