Question

The line x + y = 2 intersects curve 2(x-1)^2 - (y+1)^2 =8 at the points A and B. Find the equation of the perpendicular bisector of the line joining A and B

Answer 1

U have (-5,7) and (3,-1) -> slope 1 so ur bisector will have slope -1

Calculate mid point (x,y) and that plus the slope is your line.

Answer 2

hmm equation of line is x + y =2
\(y = 2 - x \) for point intersecting points on curve we can substitute the y value in curve's equation

\[2(x-1)^2 - (2-x+1)^2 = 8 \]
\[2(x-1)^2 - (1 -x)^2 = 8 \]
\[2(x^2 +1 -2x) - (1 + x^2 - 2x) = 8 \]
\[x^2 + 1 - 2x - 8 = 0 \]
Now you have a quadratic in x solve it you will get 2 points and similarly form a quadratic in y for getting y points
the points that you will have will be same as estudier's
for finding the the mid points \( x_{mid},y_{mid} = \frac{x_1 + x_2}{2},\frac{y_1+y_2}{2}\)
Now mid point will be (-1 , 3)

Answer 3

solve x by cross multiply right

Answer 4

Now the line the perpendicular line should pass through (-1,3) And will have minus of reciprocal of slope of the line x + y =2 (slope of line x +y = 2 is -1 ) hence, slope of perpendicular line should be 1
y = x + c
to determine c we will put the point that this line pass through we have (-1,3) as that point
3 = -1 +c
c = 4
hence, we have an equation y = x +4

Answer 5

lol i think cant you know

Answer 6

You will have to solve the quadratic through the quadratic formula or factor by group

Answer 7

in this case?

Answer 8

but I am for if there is a better method available to determine point of intersection it's lengthy one that I did

Answer 9

The quadratic equation we have is \[x^2 - 2x - 7 = 0 \]...hmm can anyone spot any errors in my solution I think I messed up somewhere

Answer 10

http://www.wolframalpha.com/input/?i=plot+x+%2B+y+%3D+2%2Cy+%3Dx%2B4%2C+2%28x-1%29^2+-+%28y%2B1%29^2+%3D8%2C+x%3D-5..20%2Cy%3D-5..20