Question Number 104309 by ajfour last updated on 20/Jul/20
Commented by ajfour last updated on 20/Jul/20
$${A},\:{B}\:{are}\:{two}\:{vertices}\:{of}\:\bigtriangleup{ABC} \\ $$$${while}\:{H}\:{is}\:{its}\:{orthocenter};\:{find} \\ $$$${coordinates}\:{of}\:{the}\:{vertex}\:{C} \\ $$$${of}\:{the}\:\bigtriangleup{ABC}. \\ $$
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)](Q104320.png)
$${BH}=\left({r}−{h}\right){i}+\left({s}−{k}\right){j} \\ $$$${AH}=\left({r}−{p}\right){i}+\left({s}−{q}\right){j} \\ $$$${BC}=\left({s}−{q}\right){i}−\left({r}−{p}\right){j} \\ $$$${AC}=\left({s}−{k}\right){i}−\left({r}−{h}\right){j} \\ $$$${OC}=\left[{h}+\lambda\left({s}−{q}\right)\right]{i}+\left[{k}−\lambda\left({r}−{p}\right)\right]{j} \\ $$$${OC}=\left[{p}+\mu\left({s}−{k}\right)\right]{i}+\left[{q}−\mu\left({r}−{h}\right)\right]{j} \\ $$$${h}+\lambda\left({s}−{q}\right)={p}+\mu\left({s}−{k}\right) \\ $$$$\Rightarrow\lambda\left({s}−{q}\right)−\mu\left({s}−{k}\right)={p}−{h}\:\:\:...\left({I}\right) \\ $$$${k}−\lambda\left({r}−{p}\right)={q}−\mu\left({r}−{h}\right) \\ $$$$\Rightarrow−\lambda\left({r}−{p}\right)+\mu\left({r}−{h}\right)={q}−{k}\:\:\:...\left({II}\right) \\ $$$$\Rightarrow\lambda=\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)} \\ $$$$ \\ $$$$\Rightarrow{x}_{{C}} ={h}+\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)}\left({s}−{q}\right) \\ $$$$\Rightarrow{y}_{{C}} ={k}−\frac{\left({r}−{h}\right)\left({p}−{h}\right)+\left({s}−{k}\right)\left({q}−{k}\right)}{\left({r}−{h}\right)\left({s}−{q}\right)−\left({s}−{k}\right)\left({r}−{p}\right)}\left({r}−{p}\right) \\ $$
Commented by ajfour last updated on 21/Jul/20
$${Slope}\:{of}\:{AB}\::\:\:\:{m}_{\mathrm{1}} =−\left(\frac{{k}−{q}}{{p}−{h}}\right) \\ $$$${eq}.\:{of}\:{HC}:\:\:{y}=\left(\frac{{p}−{h}}{{k}−{q}}\right)\left({x}−{r}\right)+{s} \\ $$$${slope}\:{of}\:{AH}:\:\:{m}_{\mathrm{2}} =−\left(\frac{{s}−{q}}{{p}−{r}}\right) \\ $$$${eq}.\:{of}\:{BC}:\:\:\:{y}=\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}−{h}\right)+{k} \\ $$$${Intersection}\:{of}\:{BC}\:{and}\:{HC}: \\ $$$$\left(\frac{{p}−{h}}{{k}−{q}}\right)\left({x}−{r}\right)+{s}\:=\:\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}−{h}\right)+{k} \\ $$$$\Rightarrow\:{x}_{{C}} =\:\frac{{r}\left(\frac{{p}−{h}}{{k}−{q}}\right)−{h}\left(\frac{{p}−{r}}{{s}−{q}}\right)+\left({k}−{s}\right)}{\left(\frac{{p}−{s}}{{k}−{q}}\right)−\left(\frac{{p}−{r}}{{s}−{q}}\right)} \\ $$$$\:\:\:\:{y}_{{C}} \:=\:\left(\frac{{p}−{r}}{{s}−{q}}\right)\left({x}_{{C}} −{h}\right)+{k}\:\:\:\bigstar \\ $$$$ \\ $$
Commented by ajfour last updated on 21/Jul/20
$${thanks}\:{Sir},\:{Very}\:{nice}\:{and}\:{perfect} \\ $$$${solution}, \\ $$