Question

In reperul cartezian xOy se considera A(-1,-2),B(1,2) si C(2,-1) . Sa se calculeze dinstanta de la punctul C la mijlocul segmentului AB.

Answer 1

mijlocul lui AB este D de coordonate Xd=Xa+Xb /2 = 0 ai Yd = Ya+Yb/2=0
D(0:0)

d (C:D) = radical din (o^2 - 2^2)^2 + [( 0^2 - (-1)^2 ]^2 = radical din -4^2 +1= radical din 17

Answer 2

$d(C,AB) \frac{|a x_{C}+b y_{C}+c| }{ \sqrt{ a^{2}+ b^{2} } } \\ \\ unde:a,b,c. sunt. coeficientii .ecuatiei .drepteiAB: \frac{x- x_{A} }{ x_{B}- x_{A} } = \frac{y- y_{A} }{ y_{B}- y_{A} } \\ \\ A( x_{A} , y_{A} )=>A(-1,-2) \\ \\ B( x_{B}, y_{B} )=>B(1,2) \\ \\ AB: \frac{x- (-1) }{ 1- (-1) } = \frac{y- (-2) }{ 2- (-2)} } \\ \\ AB: \frac{x+1 }{ 1+1 } = \frac{y+2 }{ 2+2} } \\ \\ AB: \frac{x+1 }{ 2 } = \frac{y+2 }{ 4} } \\ \\ AB:4(x+1)=2(y+2) \\ AB:4x+4=2y+4 \\ AB:4x-2y=0$
$\\ \\d(C,AB) \frac{|a x_{C}+b y_{C}+c| }{ \sqrt{ a^{2}+ b^{2} } } \\ \\ d(C,AB) \frac{|4*2+(-2)*(-1)+0| }{ \sqrt{ 4^{2}+ (-2)^{2} } } \\ \\ d(C,AB) \frac{|8+2| }{ \sqrt{ 16+ 4 } } \\ \\ d(C,AB)= \frac{|10|}{ \sqrt{20} } \\ \\d(C,AB)= \frac{10}{ \sqrt{20} } \\ \\ C( x_{C} , y_{C} )=>C(2,-1) \\ AB:4x-2y=0 \\ a=4 \\ b=-2 \\ c=0$