$Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\$

$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

4 0

4 years ago

BRAINLIEST!!!!!!!!!!!!!!!!!!!

AveGali [126]

The answer is A) reflection over x-axis

5 0

3 years ago