dane sa punkty K(1/4, - 2/8) i L(-0,25; 0,125) Oblicz dlugos wektora KL i wspolrzedne punktP dzielacego odcinek KL w stosunku 2:1 szybko prosze

Odpowiedź :

[tex]K(\frac14, -\frac28)\\L(-0,25; 0,125)\\\\Wektor KL = [x_l-x_k; y_l-y_k]\\x_l-x_k=-0,25-\frac14=-0,25-0,25=-0,5\\y_l-y_k=0,125-(-\frac28)=0,125+\frac14=0,125+0,25=0,375\\\\Wektor KL = [-0,5; 0,375]\\Wektor KL = [-\frac12; \frac38]\\[/tex]

[tex]|KL| = \sqrt{(-\frac12)^2+(\frac38)^2}\\|KL| = \sqrt{\frac14+\frac{36}{64}}\\|KL| = \sqrt{\frac{16}{64}+\frac{36}{64}}\\|KL|=\sqrt{\frac{52}{64}} = \frac{\sqrt{52}}8=\frac{2\sqrt{13}}8=\frac{\sqrt{13}}4[/tex]

[tex]P(x;y) \\\text{Wektor KP } = \frac23 \text{ wektora KL, lub wektor PL } = \frac23 \text{ wektora KL}[/tex]

[tex]KP = \frac23KL\\\\KP = [x_p-x_k; y_p-y_k][/tex]

[tex]KP = [x-\frac14; y+\frac14][/tex]

[tex]x-\frac14=\frac23*(-\frac12)\\x-\frac14=-\frac13 /+\frac14\\x=-\frac13+\frac14\\x=-\frac{4}{12}+\frac{3}{12}\\x=-\frac1{12}[/tex]

[tex]y+\frac14=\frac23*\frac38\\y+\frac14=\frac28 /-\frac14\\y=\frac14-\frac14\\y=0[/tex]

[tex]P_1=(-\frac1{12}; 0)[/tex]

[tex]PL = \frac23KL\\PL = [x_l-x_p; y_l-y_p]\\PL=[-\frac14 - x; \frac18-y][/tex]

[tex]-\frac14-x=\frac23*(-\frac12)\\-\frac14-x=-\frac13 /+\frac13\\-\frac14+\frac13-x=0 /+x\\-\frac14+\frac13=x\\-\frac{3}{12}+\frac{4}{12}=x\\\frac1{12}=x[/tex]

[tex]\frac18-y=\frac23*\frac38\\\frac18-y=\frac28 /-\frac28\\\frac18-\frac28-y=0 /+y\\\frac18-\frac28=y\\-\frac18=y[/tex]

[tex]P_2=(\frac1{12}; -\frac18)[/tex]

Odp. Dlugosc wektora KP: [tex]\frac{\sqrt{13}}4[/tex], wspolrzedne punktu P to [tex](-\frac1{12}; 0)[/tex] lub [tex](\frac1{12}; -\frac18)[/tex]