# Find the points of intersection of the curve xy=6 and the the line y=9-3x

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To find the points of intersection of the curve xy =6 and the line y = 9-3x,

Solution:

xy = 6......(1)

y = 9-3x................(2).

We eliminate y between xy = 6 and y = 9-3x by substituting y = 9-3x fot y in xy=6.

x(9-3x) = 6.

9x-3x^2= 6

9x-3x^2 -6 = 0

Multiply by (-1):

3x^2 -9x + 6 = 0

Divide by 3:

x^2-3x+2 = 0

(x-1)(x-2) = 0

x-1 = 0. Or x-2 = 0

x = 1 . Or x = 2.

To get corresponding y values, put x= 1 in xy = 6: 1*y = 6. Or y = 6.

Put x= 2 in xy =6: 2*y = 6. So y = 6/2 = 3.

Therefore (x,y) = (1,6) or (x,y) = (2,3) are the point of intersection of the curve xy = 6 and y = 9-3x.

For two curves, at the point of intersection, the x and y values are the same.

So here we have the curves: xy= 6 and y= 9-3x.

Substituting y=9-3x into the other equation,we get

x*(9-3x)=6

=> 9x- 3x^2 =6

cancelling 3

=> 3x-x^2=2

moving terms to create a quadratic equation

=> x^2- 3x +2=0

=> x^2 -2x-x+2=0

=> x(x-2)-1(x-2)=0

=>(x-1)(x-2)=0

=> x= 1 and x=2.

Now substituting this in y= 9-3x,

x=1 => y = 6

x=2 => y= 3

**So the points of intersection are ( 1, 6) and (2,3)**

xy = 6

==> y1= 6/x ........(1)

y2= 9- 3x ...........(2)

intersection point for y1 and y2 is when y1= y2

==> 9-3x = 6/x

multiply by x:

==> 9x - 3x^2 = 6

==> 3x^2 - 9x +6 = 0

Divide by 3:

==> x^2 - 3x + 2 = 0

==> (x-2)(x-1) = 0

==> x1= 1 ==> y = 6/1 = 6

==> x2= 2 ==> y2 = 6/2 = 3

**Then we have two intersction points:**

** P(1, 6) and M(2,3)**

xy=6 y=6/x (x#0)

y=9-3x

Intersection 6/x=9-3x 6=(9-3x)x

6=9x-3x^2

3x^2-9x+6=0 x^2-3x+2=o

(x-1)(x-2)=0

x=1 y= 6/1=6 (1;6)

x=2 y=6/2=3 (2;3)

Intersection(1;6) and (2;3)