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#?

Question

784 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-14T14:37:32

The answer is #option (a)#

Calculate #MA^2#, #MB^2# and #MC^2# and plug the values in the equations

#MA^2=(x-1)^2+(y+1)^2=x^2-2x+1+y^2+2y+1#

#2MA^2=2((x-1)^2+(y+1)^2)=2x^2-4x+2+2y^2+4y+2#

#MB^2=(x-0)^2+(y-3)^2=x^2+y^2-6y+9#

Therefore,

#2MA^2+MB^2=3x^2-4x+3y^2-2y+13#

#MC^2=(x-5)^2+(y-0)^2=x^2-10x+25+y^2#

#3MC^2=3(x-5)^2+3(y-0)^2=3x^2-30x+75+3y^2#

Finally,

#2MA^2+MB^2-3MC^2=3x^2-4x+3y^2-2y+13-(3x^2-30x+75+3y^2)#

#=cancel(3x^2)-4x+cancel(3y^2)-2y+13-cancel(3x^2)+30x-75-cancel(3y^2)#

#=26x-2y-62#

#=13x-y-31#

#=0#

#13x-y-31=0#

The answer is #option (a)#