Question

помогите пожалуйста 20 баллов​

Answer 1

Ответ:

$A(5;6)\ \ ,\ \ B(-7;-11)\\\\1)\ \ AB=\sqrt{(-7-5)^2+(-11-6)^2}=\sqrt{196+289}=\sqrt{485}\\\\2)\ \ AM=MB\ \ ,\ \ x_{M}=\dfrac{5-7}{2}=-1\ \ ,\ \ y_{M}=\dfrac{6-11}{2}=-2,5\\\\M(-1\ ;\ 2,5\ )\\\\3)\ \ AP:PB=3:1\ \ \to \ \ AP=3\cdot PB\ ,\ \ \lambda =3\\\\\\x_{P}=\dfrac{x_{A}+\lambda \cdot x_{B}}{1+\lambda }=\dfrac{5-3\cdot 7}{1+3}=-4\\\\\\y_{P}=\dfrac{y_{A}+\lambda \cdot y_{B}}{1+\lambda }=\dfrac{6-3\cdot 11}{1+3}=-6,75\ \ ,\qquad P(-4\ ;\, -6,75\ )$

$4)\ \ (x-5)^2+(y-6)^2=485\\\\5)\ \ R=OB=\sqrt{(-7-0)^2+(-11-0)^2}=\sqrt{49+121}=\sqrt{170}\\\\x^2+y^2=170\\\\6)\ \ y=kx+b:\ \left\{\begin{array}{l}5k+b=6\\-7k+b=-11\end{array}\right\ \ominus \ \left\{\begin{array}{l}b=6-5k\\12k=17\end{array}\right\ \ \left\{\begin{array}{l}\ b=-\dfrac{13}{12}\\k=\dfrac{17}{12}\end{array}\right\\\\y=\dfrac{17}{12}\, x-\dfrac{13}{12}\\\\\\7)\ \ k=\dfrac{17}{12}=1\dfrac{5}{12}$