Find an equation of the plane consisting of all points that are equidistant from (-1, 1, -5) and (4, -5, 0).

the required equation is 5x - 6y + 5z = 7

Step-by-step explanation:

Given that;

points : (-1, 1, -5) and (4, -5, 0).

mid point : ( [(-1+4)/2], [(1-5)/2], [(-5+0)/2]

⇒ midpoint : ( 3/2, -2, -5/2 )  

                        x₀    y₀    z₀

Direction vector n = [4-(-1)], [ -5 - 1], [ 0 - (-5)]

⇒ Direction vector n = < 5, -6, 5 >

General equation plane : n(x-x₀, y-y₀, z-z₀) = 0

so we substitute

⇒ (5, -6, 5) (x-3/2, y-(-2), z-(-5/2) ) = 0

⇒ (5, -6, 5) (x - 3/2, y + 2, z + 5/2) ) = 0

⇒ 5(x - 3/2) - 6(y + 2 ) + 5(z + 5/2) = 0

⇒ 5x - 15/2 - 6y - 12 + 5z + 25/2 = 0

⇒ 5x - 6y + 5z = 15/2 + 12 - 25/2

⇒ 5x - 6y + 5z = 7

Therefore, the required equation is 5x - 6y + 5z = 7

answer:

what is your question that you ask

step-by-step explanation:

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1