Whats the slope intercept for x-8y=-6

1 answer:

6 0

The slope-intercept form of a line is in the form y = mx + b. So basically if you want to put the equation of a line in slope-intercept form, you solve for y.
x - 8y = -6
-8y = -x - 6
y = (1/8)x + 3/4

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Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular

kozerog [31]

Hello :
let A(0,3,2) and (Δ) this line , v vector parallel to (Δ).
M∈ (Δ) : vector (AM) = t v..... t ∈ R

1 ) (Δ) parallel to the plane x + y + z = 5 : let : n an vector perpendicular
to the plane : n ⊥ v .... n(1,1,1) so : n.v =0 means : n.vector (AM) = 0
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2) (Δ) perpendicular to the line (Δ') : x = 1+t , y = 3 - t , z = 2t :
vector (u) ⊥ v .... vector(u) parallel to (Δ') and vector(u) = (1 , -1 ,1)
vector (u) ⊥ vector (AM) means :
(1)(x)+(-1)(y-3)+(2)(z -2) =0
x - y+2z - 1 = 0 ...(2)
so the system :
x+y+z - 5=0 ...(1)
x - y+2z - 1 = 0 ...(2)
(1)+(2) : 2x+3z - 6 =0
x = 3 - (3/2)z
subsct in (1) : 3 - (3/2)z +y +z - 5 =0
y = 1/2z +2
let : z=t
an parametric equations for the line (Δ) is : x = 3 - (3/2)t
 y = (1/2)t +2
z=t

verifiy :
1) (Δ) parallel to the plane x + y + z = 5 :
(-3/2 , 1/2 ,1) perpendicular to (1,1,1)
because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (Δ) perpendicular to the line (Δ') :
(-3/2 , 1/2 ,1) perpendicular to (1,-1,2)
because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0
A(0, 3, 2)∈(Δ) :
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same : t = 2

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The link down below

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Step-by-step explanation: