Wyznaczmy równanie prostej AB:

$\{\begin{array}{l}4=-3a+b\\ 7=6a+b\end{array}$    

$\{\begin{array}{l}b=4+3a\\ 7=6a+4+3a\end{array}$  

$\{\begin{array}{l}b=4+3a\\ 7=9a+4\end{array}$

$\{\begin{array}{l}b=4+3a\\ 3=9a\end{array}$

$\{\begin{array}{l}b=4+3a\\ 1=3a\end{array}$

$\{\begin{array}{l}b=4+3a\\ a=\frac{1}{3}\end{array}$

$\{\begin{array}{l}b=4+3\cdot \frac{1}{3}\\ a=\frac{1}{3}\end{array}$

$\{\begin{array}{l}b=4+1\\ a=\frac{1}{3}\end{array}$

$\{\begin{array}{l}b=5\\ a=\frac{1}{3}\end{array}$

zatem:

$AB:y=\frac{1}{3}x+5$  

zatem punkt P ma współrzędne postaci:

$P\left(x,\frac{1}{3}x+5\right),x\in \left(-3,6\right)$  

(ponieważ punkt P należy do odcinka AB)

 

Obliczmy długość odcinka AB:

$\left|AB\right|=\sqrt{{\left(-3-6\right)}^2+{\left(4-7\right)}^2}$  

$\left|AB\right|=\sqrt{{\left(-9\right)}^2+3^2}$  

$\left|AB\right|=\sqrt{81+9}$  

$\left|AB\right|=\sqrt{90}$  

$\left|AB\right|=3\sqrt{10}$  

$\left|AB\right|=\left|AP\right|+\left|PB\right|$  

zatem:

$\{\begin{array}{l}\left|AP\right|+\left|PB\right|=3\sqrt{10}\\ \frac{|AP|}{|PB|}=\frac{1}{2}\end{array}$  

$\{\begin{array}{l}\left|PB\right|=3\sqrt{10}-\left|AP\right|\\ \frac{|AP|}{3\sqrt{10}-|AP|}=\frac{1}{2}\end{array}$

$\{\begin{array}{l}\left|PB\right|=3\sqrt{10}-\left|AP\right|\\ 2\cdot \left|AP\right|=3\sqrt{10}-\left|AP\right|\end{array}$

$\{\begin{array}{l}\left|PB\right|=3\sqrt{10}-\left|AP\right|\\ 3\cdot \left|AP\right|=3\sqrt{10}\end{array}$

$\{\begin{array}{l}\left|PB\right|=3\sqrt{10}-\left|AP\right|\\ \left|AP\right|=\sqrt{10}\end{array}$

Wyznaczmy współrzędne punktu P:

$\sqrt{10}=\sqrt{{\left(x_P-\left(-3\right)\right)}^2+{\left(y_P-4\right)}^2}\mid {}^2$  

$10={\left(x_P-\left(-3\right)\right)}^2+{\left(y_P-4\right)}^2$  

$10={\left(x_P+3\right)}^2+{\left(\frac{1}{3}{x}_P+5-4\right)}^2$  

$10=x_P^2+6x_P+9+{\left(\frac{1}{3}{x}_P+1\right)}^2$  

$1=x_P^2+6x_P+\frac{1}{9}{x}_P^2+\frac{2}{3}{x}_P+1$  

$0=\frac{10}{9}{x}_P^2+\frac{20}{3}{x}_P\mid \cdot 9$  

$10x_P^2+60x_P=0\mid :10$  

$x_P^2+6x_P=0$  

$x_P\cdot \left(x_P+6\right)=0$  

$x_P=0\in \left(-3,6\right)\vee x_P=-6\notin \left(-3,6\right)$  

czyli:

$y_P=\frac{1}{3}\cdot 0+5=5$  

więc:

$P\left(0,5\right)$  

 

Odp.: C