Let #A(3,2)# and #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)#

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)#