Question Number 104309 by ajfour last updated on 20/Jul/20

Commented byajfour last updated on 20/Jul/20

A, B are two vertices of △ABC  while H is its orthocenter; find  coordinates of the vertex C  of the △ABC.

Answered by mr W last updated on 20/Jul/20

Commented bymr W last updated on 20/Jul/20

Commented bymr W last updated on 20/Jul/20

BH=(r−h)i+(s−k)j  AH=(r−p)i+(s−q)j  BC=(s−q)i−(r−p)j  AC=(s−k)i−(r−h)j  OC=[h+λ(s−q)]i+[k−λ(r−p)]j  OC=[p+μ(s−k)]i+[q−μ(r−h)]j  h+λ(s−q)=p+μ(s−k)  ⇒λ(s−q)−μ(s−k)=p−h   ...(I)  k−λ(r−p)=q−μ(r−h)  ⇒−λ(r−p)+μ(r−h)=q−k   ...(II)  ⇒λ=(((r−h)(p−h)+(s−k)(q−k))/((r−h)(s−q)−(s−k)(r−p)))    ⇒x_C =h+(((r−h)(p−h)+(s−k)(q−k))/((r−h)(s−q)−(s−k)(r−p)))(s−q)  ⇒y_C =k−(((r−h)(p−h)+(s−k)(q−k))/((r−h)(s−q)−(s−k)(r−p)))(r−p)

Commented byajfour last updated on 21/Jul/20

Slope of AB :   m_1 =−(((k−q)/(p−h)))  eq. of HC:  y=(((p−h)/(k−q)))(x−r)+s  slope of AH:  m_2 =−(((s−q)/(p−r)))  eq. of BC:   y=(((p−r)/(s−q)))(x−h)+k  Intersection of BC and HC:  (((p−h)/(k−q)))(x−r)+s = (((p−r)/(s−q)))(x−h)+k  ⇒ x_C = ((r(((p−h)/(k−q)))−h(((p−r)/(s−q)))+(k−s))/((((p−s)/(k−q)))−(((p−r)/(s−q)))))      y_C  = (((p−r)/(s−q)))(x_C −h)+k   ★

Commented byajfour last updated on 21/Jul/20

thanks Sir, Very nice and perfect  solution,