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Question-104309




Question Number 104309 by ajfour last updated on 20/Jul/20
Commented by ajfour 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.
$${A},\:{B}\:{are}\:{two}\:{vertices}\:{of}\:\bigtriangleup{ABC} \\ $$$${while}\:{H}\:{is}\:{its}\:{orthocenter};\:{find} \\ $$$${coordinates}\:{of}\:{the}\:{vertex}\:{C} \\ $$$${of}\:{the}\:\bigtriangleup{ABC}. \\ $$
Answered by mr W last updated on 20/Jul/20
Commented by mr W last updated on 20/Jul/20
Commented by mr 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)
$${BH}=\left({r}−{h}\right){i}+\left({s}−{k}\right){j} \\ $$$${AH}=\left({r}−{p}\right){i}+\left({s}−{q}\right){j} \\ $$$${BC}=\left({s}−{q}\right){i}−\left({r}−{p}\right){j} \\ $$$${AC}=\left({s}−{k}\right){i}−\left({r}−{h}\right){j} \\ $$$${OC}=\left[{h}+\lambda\left({s}−{q}\right)\right]{i}+\left[{k}−\lambda\left({r}−{p}\right)\right]{j} \\ $$$${OC}=\left[{p}+\mu\left({s}−{k}\right)\right]{i}+\left[{q}−\mu\left({r}−{h}\right)\right]{j} \\ $$$${h}+\lambda\left({s}−{q}\right)={p}+\mu\left({s}−{k}\right) \\ $$$$\Rightarrow\lambda\left({s}−{q}\right)−\mu\left({s}−{k}\right)={p}−{h}\:\:\:…\left({I}\right) \\ $$$${k}−\lambda\left({r}−{p}\right)={q}−\mu\left({r}−{h}\right) \\ $$$$\Rightarrow−\lambda\left({r}−{p}\right)+\mu\left({r}−{h}\right)={q}−{k}\:\:\:…\left({II}\right) \\ $$$$\Rightarrow\lambda=\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)} \\ $$$$ \\ $$$$\Rightarrow{x}_{{C}} ={h}+\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)}\left({s}−{q}\right) \\ $$$$\Rightarrow{y}_{{C}} ={k}−\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)}\left({r}−{p}\right) \\ $$
Commented by ajfour 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   ★
$${Slope}\:{of}\:{AB}\::\:\:\:{m}_{\mathrm{1}} =−\left(\frac{{k}−{q}}{{p}−{h}}\right) \\ $$$${eq}.\:{of}\:{HC}:\:\:{y}=\left(\frac{{p}−{h}}{{k}−{q}}\right)\left({x}−{r}\right)+{s} \\ $$$${slope}\:{of}\:{AH}:\:\:{m}_{\mathrm{2}} =−\left(\frac{{s}−{q}}{{p}−{r}}\right) \\ $$$${eq}.\:{of}\:{BC}:\:\:\:{y}=\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}−{h}\right)+{k} \\ $$$${Intersection}\:{of}\:{BC}\:{and}\:{HC}: \\ $$$$\left(\frac{{p}−{h}}{{k}−{q}}\right)\left({x}−{r}\right)+{s}\:=\:\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}−{h}\right)+{k} \\ $$$$\Rightarrow\:{x}_{{C}} =\:\frac{{r}\left(\frac{{p}−{h}}{{k}−{q}}\right)−{h}\left(\frac{{p}−{r}}{{s}−{q}}\right)+\left({k}−{s}\right)}{\left(\frac{{p}−{s}}{{k}−{q}}\right)−\left(\frac{{p}−{r}}{{s}−{q}}\right)} \\ $$$$\:\:\:\:{y}_{{C}} \:=\:\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}_{{C}} −{h}\right)+{k}\:\:\:\bigstar \\ $$$$ \\ $$
Commented by ajfour last updated on 21/Jul/20
thanks Sir, Very nice and perfect  solution,
$${thanks}\:{Sir},\:{Very}\:{nice}\:{and}\:{perfect} \\ $$$${solution}, \\ $$

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