SOLUTION: Find the equation of the line with slope -1 that is the tangent to the curve y= 1/(x-1). (5 marks: 1 mark each for a graph, for the general equation of the line, for getting the

Algebra ->  Graphs -> SOLUTION: Find the equation of the line with slope -1 that is the tangent to the curve y= 1/(x-1). (5 marks: 1 mark each for a graph, for the general equation of the line, for getting the       Log On


   

Question 322081: Find the equation of the line with slope -1 that is the tangent to the curve
y= 1/(x-1). (5 marks: 1 mark each for a graph, for the general equation of the line, for getting the quadratic, for solving for k, and for the solution).
I would really appreciate if someone could solve it for me.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
A line with a slope of -1 would have a slope-intercept form of,
y=mx%2Bb
y=-x%2Bb
Since y=1%2F%28x-1%29, then solve for b using x=0,
-0%2Bb=1%2F%280-1%29
b=1%2F-1=-1
highlight%28y=-x-1%29
.
.
.
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C1%2F%28x-1%29%2C-x-1%29
.
.
.
As you can see from graphing the equations, that only solves for one line.
You can use a symmetry argument to find the other line, y=-x%2Bb2, since the perpendicular line, y=x-1 is perpendicular to both and intersects at both intersection points.
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C1%2F%28x-1%29%2C-x-1%2Cx-1%29
.
.
.
x-1=1%2F%28x-1%29
%28x-1%29%5E2=1
x-1=0+%2B-+1
Two solutions:
x=2 and x=0
When x=2, then
-2%2Bb=1%2F%282-1%29
b=3
The other line is then,
highlight_green%28y=-x%2B3%29
.
.
.
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C1%2F%28x-1%29%2C-x-1%2Cx-1%2C-x%2B3%29
.
.
.
What's k?