Find an equation of the plane consisting of all points that are equidistant from (-3, 5, -4) and (-5, 0, 4), and having -2 as the coefficient of x

Accepted Solution

Answer:-2x-5y+8z+4.5=0Step-by-step explanation:Let (x,y,z) be the coordinates of the point lying on the needed plane. This point is equidistant from the points (-3, 5, -4) and (-5, 0, 4), so[tex]d_1=\sqrt{(x-(-3))^2+(y-5)^2+(z-(-4))^2}=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}\\ \\d_2=\sqrt{(x-(-5))^2+(y-0)^2+(z-4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\d_1=d_2\Rightarrow \sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2\\ \\x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16\\ \\-4x-10y+16z+9=0\\ \\-2x-5y+8z+4.5=0[/tex]