Question

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

Answer 1

Answer:

-2x-5y+8z+4.5=0

Step-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

$d_1=\sqrt{(x-(-3))^2+(y-5)^2+(z-(-4))^2}=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}\\ \\d_2=\sqrt{(x-(-5))^2+(y-0)^2+(z-4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\d_1=d_2\Rightarrow \sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2\\ \\x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16\\ \\-4x-10y+16z+9=0\\ \\-2x-5y+8z+4.5=0$

Answer 2

Answer:

$-2x-5y+8z+4.5=0$

Step-by-step explanation:

Let an equation of the plane

$ax+by+cz+d=0$

We have to find the equation of the plane consisting of all points that are equidistant from (-3,5,-4) and (-5,0,4).

The coefficient of x=-2

Distance formula=$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Let $d_1$ be the distance of point (x,y,z)lie on the plane and point (-3,5,-4).

$d_1=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}$

Let $d_2$ be the distance between the point (x,y,z) lie on the  plane and the point (-5,0,4).

$d_2=\sqrt{(x+5)^2+y^2+(z-4)^2}$

According to question

$d_1=d_2$

$\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}$

Squaring on both sides

$(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2$

$x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16$

$x^2+6x+50+y^2-10y+z^2+8z-x^2-10x-z^2-y^2+8z-41=0$

$-4x-10y+16z+9=0$

Divided the equation by 2

$-2x-5y+8z+4.5=0$