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Alina [70]
3 years ago
6

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 th

e coefficient of x
Mathematics
2 answers:
stiks02 [169]3 years ago
8 0

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

Firdavs [7]3 years ago
5 0

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

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