Coordinate Geometry

Question

A straight line through the point (0,-3) intersects the curve x^2+y^2-27x+41=0 at (2,3).Calculate the coordinates of the point at which the line again meets the curve.

Ans (2.5,4.5)

Answer 1

First we need the equation of the line through the points (0,-3), (2,3)

The slope is rise/run    m = (3-(-3))/(2-0) = 6/2 = 3

Using slope intercept form y=mx+b where m is the slope and b is the y intercept we have...

The line connecting our two points will have form  y = 3x +b

We already have the y-intercept which is (0,-3) so b = -3

 

The line connecting the two points is   y=3x-3

 

This line connects with the curve x²+y²-27x+41 at (2,3) and one other point.

 

We want to solve the system of two equations.

x²+y²-27x+41=0   eqn 1

y=3x-3             eqn 2

....  Substitute 3x-3 in place of y in the equation x²+y²-27x+41

 

x²+ (3x-3)²-27x+41=0

x²+(3x-3)(3x-3)-27x+41=0

x²+9x²-9x-9x+9-27x+41=0

10x²-45x+50=0

We can factor out a 5

5(2x²-9x+10)=0

2x²-9x+10 = 0

 

We can factor this by grouping

 

2x²-4x-5x+10 = 0

2x(x-2)-5(x-2)=0

(2x-5)(x-2)=0

 

Either 2x-5=0      or     x-2 =0

         2x=5

          x=5/2

          x=2.5        or     x=2

 

We already have the point of intersection where x=2 given as (2,3)

We need the point of intersection where x=2.5

 

Let's use the easier equation...    y=3x-3

                                                y=3(2.5)-3

                                                y=7.5-3

                                                y=4.5

 

The 2nd point of intersection is at point (2.5, 4.5)