# We have points A(1,-1);B(0,3);C(5,0).Geometric place of points M(x,y) for which 2MA^2+MB^2-3MC^2=0 it is: a)right of equation 13x-y-31=0 b)right of equation y=0 c)right of equation 13x-y-30=0 d)circle of equation x^2+y^2-3x+2y-16=o?

Jul 14, 2018

The answer is $o p t i o n \left(a\right)$

#### Explanation:

Calculate $M {A}^{2}$, $M {B}^{2}$ and $M {C}^{2}$ and plug the values in the equations

$M {A}^{2} = {\left(x - 1\right)}^{2} + {\left(y + 1\right)}^{2} = {x}^{2} - 2 x + 1 + {y}^{2} + 2 y + 1$

$2 M {A}^{2} = 2 \left({\left(x - 1\right)}^{2} + {\left(y + 1\right)}^{2}\right) = 2 {x}^{2} - 4 x + 2 + 2 {y}^{2} + 4 y + 2$

$M {B}^{2} = {\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = {x}^{2} + {y}^{2} - 6 y + 9$

Therefore,

$2 M {A}^{2} + M {B}^{2} = 3 {x}^{2} - 4 x + 3 {y}^{2} - 2 y + 13$

$M {C}^{2} = {\left(x - 5\right)}^{2} + {\left(y - 0\right)}^{2} = {x}^{2} - 10 x + 25 + {y}^{2}$

$3 M {C}^{2} = 3 {\left(x - 5\right)}^{2} + 3 {\left(y - 0\right)}^{2} = 3 {x}^{2} - 30 x + 75 + 3 {y}^{2}$

Finally,

$2 M {A}^{2} + M {B}^{2} - 3 M {C}^{2} = 3 {x}^{2} - 4 x + 3 {y}^{2} - 2 y + 13 - \left(3 {x}^{2} - 30 x + 75 + 3 {y}^{2}\right)$

$= \cancel{3 {x}^{2}} - 4 x + \cancel{3 {y}^{2}} - 2 y + 13 - \cancel{3 {x}^{2}} + 30 x - 75 - \cancel{3 {y}^{2}}$

$= 26 x - 2 y - 62$

$= 13 x - y - 31$

$= 0$

$13 x - y - 31 = 0$

The answer is $o p t i o n \left(a\right)$