Let A(3,2)A(3,2) and B(5,1)B(5,1). ABP is an equilateral triangle constructed on the side of AB remote from origin then the orthocentre of triangle ABP is?

2 Answers
Oct 6, 2017

((24+sqrt3)/6,(9+2sqrt3)/6)(24+36,9+236)

Explanation:

enter image source here
Given A(3,2), B(5,1), and Delta ABP (P remote from the origin) is an equilateral triangle,
=> AB=BP=PA=sqrt((5-3)^2+(1-2)^2)=sqrt5
Let M be the midpoint of AB,
=> M(x_m,y_m)=((3+5)/2,(2+1)/2)=(4,3/2)
slope of AB=(1-2)/(5-3)=-1/2
As MP is perpendicular to AB,
=> slope of MP=2
=> tantheta=2,
=> costheta=1/sqrt5, sintheta=2/sqrt5
MP="AP"sin60=sqrt5*sqrt3/2=sqrt15/2
=> P(x_p,y_p)=(x_m+MPcostheta, " "y_m+"MP"sintheta)
=(4+(sqrt15/2xx1/sqrt5), " "3/2+(sqrt15/2xx2/sqrt5))
=> (4+sqrt3/2, " "3/2+sqrt3)
Recall that color(red)"the orthocenter and the centroid of an equilateral triangle" are the same point, and a triangle with vertices at (x_1,y_1), (x_2,y_2), (x_3,y_3) has centroid at
((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3)

=> orthocenter of DeltaABP is
O(x,y)=((3+5+4+sqrt3/2)/3, " "(2+1+3/2+sqrt3)/3)
=((24+sqrt3)/6," "(9+2sqrt3)/6)

Oct 6, 2017

((24+sqrt3)/6, " "(9+2sqrt3)/6)

Explanation:

Solution 2)
enter image source here
Given A(3,2), B(5,1), and Delta ABP (P remote from the origin) is an equilateral triangle,
=> AB=BP=PA=sqrt((5-3)^2+(1-2)^2)=sqrt5
Let M be the midpoint of AB,
=> M(x_m,y_m)=((3+5)/2,(2+1)/2)=(4,3/2)
slope of AB=(1-2)/(5-3)=-1/2
As MP is perpendicular to AB,
=> slope of MP=2
=> tantheta=2,
=> costheta=1/sqrt5, sintheta=2/sqrt5
MP="AP"sin60=sqrt5*sqrt3/2=sqrt15/2

Recall that in an equilateral triangle, the orthocenter and the centroid coincide, and that the centroid of a triangle divides the medians into segment with a 2:1 ratio,
=> MO=1/3MP
=> O(x,y)=(x_m+1/3MPcostheta, " "y_m+1/3"MP"sintheta)
=(4+(1/3xxsqrt15/2xx1/sqrt5), " "3/2+(1/3xxsqrt15/2xx2/sqrt5))
=> ((24+sqrt3)/6, " "(9+2sqrt3)/6)