Ответ:
Уравнение прямой проходящей через две точки:
[tex]\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}[/tex]
[tex]A(5;-3)\ ,\ B(-1;-2)\\\\\dfrac{x-5}{-1-5}=\dfrac{y+3}{-2+3}\ \ \ \Rightarrow \ \ \ \ \dfrac{x-5}{-6}=\dfrac{y+3}{1}\ \ ,\\\\\\x-5=-6\, (y+3)\\\\x-5=-6y-18\\\\AB:\ \ \underline {x+6y+13=0}\ \ \ \Rightarrow \ \ \ \ 6y=-x-13\ \ .\ \ \ \underline{y=-\dfrac{x}{6}-\dfrac{13}{6}\ }[/tex]
Объяснение:
[tex]A(5;-3)\ \ \ \ \ \ B(-1;-2)\ \ \ \ \ y=kx+b\ ?\\\left \{ {{-3=k*5+b} \atop {-2=k*(-1)+b}} \right.\ \ \ \ \ \left \{ {{5k+b=-3} \atop {-k+b=-2\ |*5}} \right. \ \ \ \ \left \{ {{5k+b=-3} \atop {-5k+5b=-10}} \right. .[/tex]
Суммируем эти уравнения:
[tex]6b=-13\ |:6\\b=-\frac{13}{6} \\-k+(-\frac{13}{6} )=-2\ |*(-1)\\k+2\frac{1}{6} =2\\k=-\frac{1}{6}.\ \ \ \ \ \Rightarrow\\ y=-\frac{1}{6}x+(-\frac{13}{6})=-\frac{x+13}{6} .\\OTBET: \ y=-\frac{x+13}{6} .[/tex]
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Ответ:
Уравнение прямой проходящей через две точки:
[tex]\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}[/tex]
[tex]A(5;-3)\ ,\ B(-1;-2)\\\\\dfrac{x-5}{-1-5}=\dfrac{y+3}{-2+3}\ \ \ \Rightarrow \ \ \ \ \dfrac{x-5}{-6}=\dfrac{y+3}{1}\ \ ,\\\\\\x-5=-6\, (y+3)\\\\x-5=-6y-18\\\\AB:\ \ \underline {x+6y+13=0}\ \ \ \Rightarrow \ \ \ \ 6y=-x-13\ \ .\ \ \ \underline{y=-\dfrac{x}{6}-\dfrac{13}{6}\ }[/tex]
Объяснение:
[tex]A(5;-3)\ \ \ \ \ \ B(-1;-2)\ \ \ \ \ y=kx+b\ ?\\\left \{ {{-3=k*5+b} \atop {-2=k*(-1)+b}} \right.\ \ \ \ \ \left \{ {{5k+b=-3} \atop {-k+b=-2\ |*5}} \right. \ \ \ \ \left \{ {{5k+b=-3} \atop {-5k+5b=-10}} \right. .[/tex]
Суммируем эти уравнения:
[tex]6b=-13\ |:6\\b=-\frac{13}{6} \\-k+(-\frac{13}{6} )=-2\ |*(-1)\\k+2\frac{1}{6} =2\\k=-\frac{1}{6}.\ \ \ \ \ \Rightarrow\\ y=-\frac{1}{6}x+(-\frac{13}{6})=-\frac{x+13}{6} .\\OTBET: \ y=-\frac{x+13}{6} .[/tex]