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A is the point (-3, 6). Let (𝑥, 𝑦) be any point on the line 𝑦 = 2𝑥.

a) Write an equation in terms of 𝑥 for the distance between (𝑥, 𝑦) and A(-3, 6).

b) Find the coordinates of the two points, B and C, on the line 𝑦 = 2𝑥 and which are a distance of 6 from (-3, 6).

c) Find the equation of 𝑙1 that is perpendicular to 𝑦 = 2𝑥 and goes through the point (-3, 6) in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0, 𝑤ℎ𝑒𝑟𝑒 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠.

please explain each steps thanks!!!!

 Oct 29, 2018
edited by YEEEEEET  Oct 29, 2018
 #1
avatar+98044 
+1

A is the point (-3, 6). Let (𝑥, 𝑦) be any point on the line 𝑦 = 2𝑥.

a) Write an equation in terms of 𝑥 for the distance between (𝑥, 𝑦) and A(-3, 6).

b) Find the coordinates of the two points, B and C, on the line 𝑦 = 2𝑥 and which are a distance of 6 from (-3, 6).

c) Find the equation of 𝑙1 that is perpendicular to 𝑦 = 2𝑥 and goes through the point (-3, 6) in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0, 𝑤ℎ𝑒𝑟𝑒 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠.

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

 

 

cool cool cool

 Oct 29, 2018

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