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# A is the point (-3, 6). Let (π₯, π¦) be any point on the line π¦ = 2π₯.

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A is the point (-3, 6). Let (π₯, π¦) be any point on the line π¦ = 2π₯.

a) Write an equation in terms of π₯ for the distance between (π₯, π¦) and A(-3, 6).

b) Find the coordinates of the two points, B and C, on the line π¦ = 2π₯ and which are a distance of 6 from (-3, 6).

c) Find the equation of π1 that is perpendicular to π¦ = 2π₯ and goes through the point (-3, 6) in the form ππ₯ + ππ¦ + π = 0, π€βπππ π, π πππ π πππ πππ‘πππππ .

Oct 29, 2018
edited by YEEEEEET  Oct 29, 2018

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A is the point (-3, 6). Let (π₯, π¦) be any point on the line π¦ = 2π₯.

a) Write an equation in terms of π₯ for the distance between (π₯, π¦) and A(-3, 6).

b) Find the coordinates of the two points, B and C, on the line π¦ = 2π₯ and which are a distance of 6 from (-3, 6).

c) Find the equation of π1 that is perpendicular to π¦ = 2π₯ and goes through the point (-3, 6) in the form ππ₯ + ππ¦ + π = 0, π€βπππ π, π πππ π πππ πππ‘πππππ .

a)

distance  = sqrt  [ ( x + 3)^2  + ( y - 6)^2]

b)

Let  y   = 2x  and we want to solve this

6  = sqrt [ ( x + 3)^2  + (2x - 6)^2 ]     square both sides

36  =  ( x + 6x + 9)  + ( 4x^2 - 24x + 36 )   simplify

36  = 5x^2 - 18x + 45    rearrange as

5x^2  - 18x  + 9  = 0        factor

(5x - 3) ( x  - 3)   = 0

Set both factors to 0  and solve for x and we get that

x  = 3/5  or   x  =  3

And the associated y coordinates are  y  = 2(3/5)  = 6/5

And   y  = 2(3)  =  6

So...the two points are    ( 3/5, 6/5)   and  ( 3, 6)

c )  Perpendicular to y  = 2x   and  passing through (-3, 6)

The slope of this line is   (-1/2)

So....the equation is

y  = (-1/2) ( x + 3) + 6

y = (-1/2)x - 3/2 + 12/2

y = (-1/2)x + 9/2

Oct 29, 2018