Can someone help with #12 please
1 answer:
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Answer:
D) y = -4/5x – 47/10
Step-by-step explanation:
Step 1. Find the <em>midpoint of the segmen</em>t.
The two end points are (-6, -4) and (-2, 1).
The midpoint is at the average of the coordinates.
(xₚ, yₚ) = ((x₁ + x₂)/2, (y₁ + y₂)/2)
(xₚ, yₚ) = ((-6 - 2)/2, (-4 + 1)/2)
(xₚ, yₚ) = (-8/2, -3/2)
(xₚ, yₚ) = (-4, -3/2)
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Step 2. Find the <em>slope (m₁) of the segment</em>
m₁ = (y₂ - y₁)/(x₂ - x₁)
m₁ = (1 - (-4))/(-2 - (-6))
m₁ = (1 + 4)/(-2 + 6)
m₁ = 5/4
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Step 3. Find the <em>slope (m₂) of the perpendicular bisector </em>
m₂ = -1/m₁
m₂ = -4/5
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Step 4. Find the <em>intercept of the perpendicular bisector</em>
y = mx + b
y = -(4/5)x + b
The line passes through (-4, -3/2).
-3/2 = -(4/5)(-4) + b
-3/2 = 16/5 + b Multiply each side by 10
-15 = 32 + 10 b Subtract 32 from each side
-47 = 10b Divide each side by 10
b = -47/10
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Step 5. Write the <em>equation for the perpendicular bisector</em>
y = -4/5x – 47/10
The graph shows the midpoint of your segment at (-4, -3/2) and the perpendicular bisector passing through the midpoint and (0, -47/10).
Answer:
x^2+y^2+z^2-10x-6y-10z +50 =0
Step-by-step explanation:
Given that a sphere is contained in the first octant
Centre of the sphere is given as (5,3,5)
Since this is contained only in the first octant radius should be at most sufficient to touch any one of the three coordinate planes
When it touches we can get the maximum sphere
We find that y coordinate is the minimum of 3 thus radius can be atmost 3 so that then only it can touch y =0 plane i.e. zx plane without crossing to go to the other octants.
Hence radius =3
Equation of the sphere would be