Question

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?

Answer 1

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

Explanation:

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)`

Answer 2

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

Explanation:

Solution 2)

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)`