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−ya=yb−yaxb−xa⋅(x−xa)
y-3=(4-3)/(7-1)(x-1)y−3=4−37−1(x−1)
y-3=1/6*(x-1)y−3=16⋅(x−1)
6y-18=x-16y−18=x−1
x-6y+17=0" "x−6y+17=0 first equation
Solve for the length ll using "distance from line to a point" formula:
from the first equation a=1a=1 and b=-6b=−6 and c=17c=17
l=(ax_c+by_c+c)/(+-sqrt(a^2+b^2))=(1*5-6*8+17)/(+-sqrt(1^2+(-6)^2)l=axc+byc+c±√a2+b2=1⋅5−6⋅8+17±√12+(−6)2
l=(5-48+17)/(+-sqrt(37))=(-26)/(-sqrt37)l=5−48+17±√37=−26−√37
l=4.27437l=4.27437
We need another equation to solve for the endpoints
Use point C(x_c, y_c)=C(5, 8)C(xc,yc)=C(5,8) and slope m=-6m=−6. This is perpendicular to side c or line segment AB.
y-y_c=m(x-x_c)y−yc=m(x−xc)
y-8=-6(x-5)y−8=−6(x−5)
y-8=-6x+30y−8=−6x+30
6x+y=386x+y=38
simultaneous solution of the two equations to solve for the other endpoint
x-6y+17=0" "x−6y+17=0 first equation
6x+y=38" "6x+y=38 second equation
x=6y-17x=6y−17
x=6(38-6x)-17x=6(38−6x)−17
x=228-36x-17x=228−36x−17
37x=21137x=211
x=211/37x=21137
x=5.70x=5.70
y=38-6xy=38−6x
y=38-6(211/37)=(1406-1266)/37=140/37y=38−6(21137)=1406−126637=14037
y=140/37y=14037
y=3.78y=3.78
the other endpoint is (5.70, 3.78)(5.70,3.78)
God bless...I hope the explanation is useful.
