Question

A straight line through the point (0,-3) intersects the curve y^2 + x^2 -27x +41 =0 at (2,3). calculate the coordinates of the point at which the line again meet the curve. can someone please explain this to me?

can someone please explain this to me?

Answer 1

The coordinates where the line meets the curve again is :
`(5/2, 9/2)`

Explanation:

Let the line be `y = m*x + b` that passes through the points `(0,-3)` and `(2,3).` Use these points to find `m,b`
`-3 = m(0) + b`
`3 = 2*m + b`
Solve simultaneuosly to get `m = 3, b =-3`
The solve these two equations
`y = 3x -3`
`y^2 + x^2 -27x + 41 =0`
by susbstitution for `x`
`(3x-3)^2 + x^2 -27*x +41 =0.`
This gives
`x=2, x= 5/2`
Now, sub in the values of `x` into `y=3x-3`
`y=3(2)-3 =3`, which gives coordinate `(2,3)` GIVEN ALREADY
`y = 3(5/2) -3 = 9/2` which gives the coordinate `(5/2,9/2)`, WHICH IS THE COORDINATE WE ARE LOOKING FOR.