A triangle has corners A, B, and C located at #(1 ,3 )#, #(7 ,4 )#, and #(5 , 8 )#, respectively. What are the endpoints and length of the altitude going through corner C?

1 Answer

Endpoints are #C(5, 8)# and #(211/37, 140/37)#
Length #l=4.27437#

Explanation:

We need the equation of the line passing thru A(1, 3) and B(7, 4) by the two-point form

#y-y_a=(y_b-y_a)/(x_b-x_a)*(x-x_a)#

#y-3=(4-3)/(7-1)(x-1)#

#y-3=1/6*(x-1)#

#6y-18=x-1#

#x-6y+17=0" "#first equation

Solve for the length #l# using "distance from line to a point" formula:
from the first equation #a=1# and #b=-6# and #c=17#

#l=(ax_c+by_c+c)/(+-sqrt(a^2+b^2))=(1*5-6*8+17)/(+-sqrt(1^2+(-6)^2)#
#l=(5-48+17)/(+-sqrt(37))=(-26)/(-sqrt37)#
#l=4.27437#

We need another equation to solve for the endpoints
Use point #C(x_c, y_c)=C(5, 8)# and slope #m=-6#. This is perpendicular to side c or line segment AB.

#y-y_c=m(x-x_c)#
#y-8=-6(x-5)#
#y-8=-6x+30#
#6x+y=38#

simultaneous solution of the two equations to solve for the other endpoint
#x-6y+17=0" "# first equation
#6x+y=38" "#second equation

#x=6y-17#
#x=6(38-6x)-17#
#x=228-36x-17#
#37x=211#
#x=211/37#

#x=5.70#

#y=38-6x#
#y=38-6(211/37)=(1406-1266)/37=140/37#
#y=140/37#

#y=3.78#

the other endpoint is #(5.70, 3.78)#

God bless...I hope the explanation is useful.