Geometria analityczna. Funkcja liniowa.
Znajdź wzór funkcji liniowej, której wykres przechodzi przez punkty:
[tex]a)\ A(-15,\ 2),\ B(-10,-20)\to f(x)=-\dfrac{22}{5}x-64\\\\b)\ A(-6,-3),\ B(-4,\ 2)\to f(x)=\dfrac{5}{2}x+12\\\\c)\ A(-3,\ 4),\ B\left(\dfrac{1}{2},\ 2\right)\to f(x)=-\dfrac{4}{7}x+\dfrac{16}{7}\\\\d)\ A\left(\dfrac{3}{2},\ 5\right),\ B(8,-3)\to f(x)=-\dfrac{16}{13}x+\dfrac{89}{13}\\\\e)\ A\left(\dfrac{5}{4},\ \dfrac{1}{2}\right),\ B\left(\dfrac{11}{3},\ \dfrac{4}{3}\right)\to f(x)=\dfrac{10}{29}x+\dfrac{2}{29}[/tex]
Rozwiązanie:
Skorzystamy ze wzoru na równanie prostej w postaci kierunkowej
(y = ax + b) przechodzącej przez dwa punkty:
[tex]A(x_A,\ y_A),\ B(x_B,\ y_B)\\\\y=\dfrac{y_B-y_A}{x_B-x_A}(x-x_A)+y_A[/tex]
Podstawiamy współrzędne punktów:
[tex]a)\ A(-15,\ 2)\ B(-10,-20)\\\\y=\dfrac{-20-2}{-10-(-15)}(x-(-15))+2\\\\y=\dfrac{-22}{5}(x+15)+2\\\\y=-\dfrac{22}{5}x-66+2\\\\\huge\boxed{y=-\dfrac{22}{5}x-64}[/tex]
[tex]b)\ A(-6,-3),\ B(-4,\ 2)\\\\y=\dfrac{2-(-3)}{-4-(-6)}(x-(-6))-3\\\\y=\dfrac{5}{2}(x+6)-3\\\\y=\dfrac{5}{2}x+15-3\\\\\huge\boxed{y=\dfrac{5}{2}x+12}[/tex]
[tex]c)\ A(-3,\ 4),\ B(-4,\ 2)\\\\y=\dfrac{2-4}{\frac{1}{2}-(-3)}(x-(-3))+4\\\\y=\dfrac{-2}{3\frac{1}{2}}(x+3)+4\\\\y=-\dfrac{2}{\frac{7}{2}}(x+3)+4\\\\y=-2\cdot\dfrac{2}{7}(x+3)+4\\\\y=-\dfrac{4}{7}(x+3)+4\\\\y=-\dfrac{4}{7}x-\dfrac{12}{7}+\dfrac{28}{7}\\\\\huge\boxed{y=-\dfrac{4}{7}x+\dfrac{16}{7}}}[/tex]
[tex]d)\ A\left(\dfrac{3}{2},\ 5\right),\ B(8,-3)\\\\y=\dfrac{-3-5}{8-\frac{3}{2}}\left(x-\dfrac{3}{2}\right)+5\\\\y=\dfrac{-8}{\frac{16}{2}-\frac{3}{2}}\left(x-\dfrac{3}{2}\right)+5\\\\y=-\dfrac{8}{\frac{13}{2}}\left(x-\dfrac{3}{2}\right)+5\\\\y=-8\cdot\dfrac{2}{13}\left(x-\dfrac{3}{2}\right)+5\\\\y=-\dfrac{16}{13}\left(x-\dfrac{3}{2}\right)+5\\\\y=-\dfrac{16}{13}x+\dfrac{24}{13}+\dfrac{65}{13}\\\\\huge\boxed{y=-\dfrac{16}{13}x+\dfrac{89}{13}}[/tex]
[tex]e)\ A\left(\dfrac{5}{4},\ \dfrac{1}{2}\right),\ B\left(\dfrac{11}{3},\ \dfrac{4}{3}\right)\\\\y=\dfrac{\frac{4}{3}-\frac{1}{2}}{\frac{11}{3}-\frac{5}{4}}\left(x-\dfrac{5}{4}\right)+\dfrac{1}{2}\\\\y=\dfrac{\frac{8}{6}-\frac{3}{6}}{\frac{44}{12}-\frac{15}{12}}\left(x-\dfrac{5}{4}\right)+\dfrac{1}{2}\\\\y=\dfrac{\frac{5}{6}}{\frac{29}{12}}\left(x-\dfrac{5}{4}\right)+\dfrac{1}{2}\\\\y=\dfrac{5}{6\!\!\!\!\diagup_1}\cdot\dfrac{12\!\!\!\!\!\diagup^2}{29}\left(x-\dfrac{5}{4}\right)+\dfrac{1}{2}\\\\y=\dfrac{10}{29}\left(x-\dfrac{5}{4}\right)+\dfrac{1}{2}\\\\y=\dfrac{10}{29}x-\dfrac{25}{58}+\dfrac{29}{58}\\\\y=\dfrac{10}{29}x+\dfrac{4}{58}\\\\\huge\boxed{y=\dfrac{10}{29}x+\dfrac{2}{29}}[/tex]
Wykresem funkcji liniowej f(x) = ax + b jest prosta o równaniu y = ax + b.
