+0  
 
0
45
1
avatar

Two lines pass through the point (–6, 7) and each is a distance of 2 from the origin at their closest to the origin. What is the sum of the slopes of these lines?

 Jun 22, 2021
 #1
avatar+120930 
+1

See the image  here

 

 

The two lines  will be  tangent to the  circle with the equation  x^2  +  y^2  =  4

 

The radius of this circle, and  each  tangent  line  will  form legs of two congruent right triangles  with AB  the  hypotenuse  of both

A =  (0,0)    B  = (-6,7)   

 

AC =  AD  =   2       AB  =sqrt ( 6^2  + 7^2)  = sqrt (85)

 

So 

 AB^2   -  AC^2  =  BC^2

85  -  4   =   BC^2

81  =  BC

9 = BC = BD

 

If  we  draw  a circle with radius  = 9   centered  at  (-6,7)    we  have

(x + 6)^2   +  (y - 7)^2  =   81

x^2  + 12x  +  36 +  y^2  - 14y +  49 = 81

X^2 + y^2  + 12x  - 14y   =  -4

      4  +   12x  - 14y =  -4

          12x  - 14y  =  -8

            6x  - 7y  =  -4

            7y =  6x +  4

              y=   (6x + 4)  / 7  

 

Sub the  last equation into  x^2 + y^2  = 4   and we  will  find the  x intercepts  of the tangent lines  with the  circle x^2 + y^2 =  4

 

x^2 +  ( 6x + 4)^2 /49   =  4

Solving  this  produces  x = -30/17    and  x =  6/5

And  when x = -30/17,     y = -16/17  =  C

And when x =  6/5 ,     y =  8/5     =  D

 

Slope  of   BC  =   (7 - -16/17)  / ( -6 - -30/17)    =  (7 + 16/17)   /( -6 + 30/17)   =  -15/8

Slope  of BD =   (7 - 8/5)  /( -6 - 6/5)   = -3/4  =   - 6/8

 

Sum  of the  slopes =  -15/8  - 6/8  =   - 21   /  8

 

 

cool cool cool

 Jun 22, 2021
edited by CPhill  Jun 22, 2021

6 Online Users