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Question 1180903: Given triangle ABC with vertices A(-1, 7), B(-3, -7), and C(8, 4), let AP, BQ, and CR be the perpendiculars dropped from A, B, and C to their opposite sides of triangle ABC.
Find the equations of perpendiculars AP, BQ, and CR.
Note: Show full solution pls. Ty!
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! A(-1, 7), B(-3, -7)
slope of BC
x1 y1 x2 y2
-3 -7 8 4
slope m = (y2-y1)/(x2-x1)
( 4 - -7 )/( 8 - -3 )
( 11 / 11 )
m= 1
AP is perpendicular to BC
slope of AP will be negative reciprocal of slope of BC = -1
The line passes through A (-1,7)
Find b by plugging the values of m & the point in -1
y=mx+b
7 = 1.00 + b
b= 6.00
m= -1
The required equation is y = -1 x + 6 (equation of AP)
equation of BQ
x1 y1 x2 y2
-1 7 8 4
slope m = (y2-y1)/(x2-x1)
( 4 - 7 )/( 8 - -1 )
( -3 / 9 )
m= -1/3
BQ is perpendicular to AC
slope of BQ will be negative reciprocal of slope of AC = 3
m= 3 ,point ( -3 , -7 )
Find b by plugging the values of m & the point in 3
y=mx+b
-7 = -9.00 + b
b= 2.00
m= 3
The required equation is y =3 x + 2 Equation of BQ
You continue for the third
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