SOLUTION: A line through(2 , 1)meets the curve x squared - 2x - y=3 at A(-2 , 5) and at B.Find the coordinates of B.

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Question 618104: A line through(2 , 1)meets the curve x squared - 2x - y=3 at A(-2 , 5) and at B.Find the coordinates of B.
Found 2 solutions by lwsshak3, ewatrrr:
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
A line through(2 , 1)meets the curve x squared - 2x - y=3 at A(-2 , 5) and at B.Find the coordinates of B.
**
Equation of line: y=mx+b, m=slope, b=y-intercept
m=∆y/∆x=(5-1)/(-2-2)=-4/4=-1
equation: y=-x+b
solving for b using coordinates of one of given points ((2,1)
1=-2+b
b=3
equation of line: y=-x+3
..
Equation of curve: x^2-2x-y=3
Substitute equation of line for y
x^2-2x-(-x+3)=3
x^2-2x+x-3=3
x^2-x-6=0
(x-3)(x+2)=0
..
x+2=0
x=-2
y=-x+3=2+3=5
This confirms given point of intersection at A: (-2,5)
..
x-3=0
x=3
y=-x+3=3-3=0
coordinates of B, point of intersection: (3,0)
see graph below as a visual check:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi, 

A line through(2 , 1)meets the curve x^2- 2x - y=3 at A(-2 , 5) and at B

the line passing through (2,1) and (-2,5)

(2,1) and 

(-2,5)  m = -4/4 = -1

 y = -x + b

 1 = -2 + b     ||Using PT(2,1) to solve for b

  3 = b

  = -x + 3   AND  x^2- 2x - y=3 or  = x^2- 2x - 3

  x^2-2x-3 = -x + 3

  x^2-x - 6 = 0

 (x-3)(x+2) = 0

  (x+2) = 0  We have A(-2 , 5)

 (x-3) = 0    B(3,0)      |||  = -x + 3 



 

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