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 π1 that is perpendicular to π¦ = 2π₯ and goes through the point (-3, 6) in the form ππ₯ + ππ¦ + π = 0, π€βπππ π, π πππ π πππ πππ‘πππππ .

please explain each steps thanks!!!!

YEEEEEET Oct 29, 2018

#1**+1 **

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 π1 that is perpendicular to π¦ = 2π₯ and goes through the point (-3, 6) in the form ππ₯ + ππ¦ + π = 0, π€βπππ π, π πππ π πππ πππ‘πππππ .

please explain each steps thanks!!!!

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

CPhill Oct 29, 2018