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.
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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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