Question

mng giúp em c4 với

Answer 1

a.

\(d\left(A;d\right)=\dfrac{\left|4.\left(-3\right)-3.5+8\right|}{\sqrt{4^2+\left(-3\right)^2}}=-\dfrac{19}{5}\)

b. 

Do \(\Delta\perp d\) nên \(\Delta\) nhận (3;4) là 1 vtpt

Phương trình \(\Delta\) có dạng: \(3x+4y+c=0\)

\(d\left(A;\Delta\right)=2\Leftrightarrow\dfrac{\left|-3.3+4.5+c\right|}{\sqrt{3^2+4^2}}=2\)

\(\Leftrightarrow\left|c+11\right|=10\Rightarrow\left[{}\begin{matrix}c=-21\\c=-1\end{matrix}\right.\)

Có 2 đường thẳng thỏa mãn: \(\left[{}\begin{matrix}3x+4y-1=0\\3x+4y-21=0\end{matrix}\right.\)

c.

Do \(M\in\left(a\right)\) nên tọa độ có dạng: \(M\left(2m+1;m\right)\)

\(d\left(M;d\right)=\dfrac{\left|4\left(2m+1\right)-3m+8\right|}{\sqrt{4^2+\left(-3\right)^2}}=4\)

\(\Leftrightarrow\left|5m+12\right|=20\Rightarrow\left[{}\begin{matrix}m=\dfrac{8}{5}\\m=-\dfrac{32}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}M\left(\dfrac{21}{5};\dfrac{8}{5}\right)\\M\left(-\dfrac{59}{5};-\dfrac{32}{5}\right)\end{matrix}\right.\)