SOLUTION: The equation of a curve xy = 12 and the equation of a line L is 2x + y = k, where k is a constant. i) In the case where k = 11, find the coordinates of the points of intersction

Algebra ->  Equations -> SOLUTION: The equation of a curve xy = 12 and the equation of a line L is 2x + y = k, where k is a constant. i) In the case where k = 11, find the coordinates of the points of intersction       Log On


   

Question 1200186: The equation of a curve xy = 12 and the equation of a line L is 2x + y = k, where k is
a constant.
i) In the case where k = 11, find the coordinates of the points of intersction of L
and the curve.
ii) Find the set of values of & for which I does not intersect the curve.
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

I'll answer part (i) only.
Part (ii) seems a bit garbled so you'll need to revise.


Part (i)

Plug k = 11 into the 2nd equation and isolate y.
2x+y = k
2x+y = 11
y = -2x+11

Plug this into the other equation.
xy = 12
x(-2x+11) = 12
-2x^2+11x = 12
-2x^2+11x-12 = 0

Turn to the quadratic formula.
We'll plug in
a = -2
b = 11
c = -12
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-11%2B-sqrt%28%2811%29%5E2-4%28-2%29%28-12%29%29%29%2F%282%28-2%29%29

x+=+%28-11%2B-sqrt%28121+-+96%29%29%2F%28-4%29

x+=+%28-11%2B-sqrt%2825%29%29%2F%28-4%29

x+=+%28-11%2B-++5%29%2F%28-4%29

x+=+%28-11%2B5%29%2F%28-4%29 or x+=+%28-11-5%29%2F%28-4%29

x+=+%28-6%29%2F%28-4%29 or x+=+%28-16%29%2F%28-4%29

x+=+3%2F2 or x+=+4

x+=+1.5 or x+=+4

If x = 1.5, then,
y = -2x+11
y = -2*1.5+11
y = -3+11
y = 8
Making (1.5, 8) one point of intersection.

If x = 4, then,
y = -2x+11
y = -2*4+11
y = -8+11
y = 3
Making (4, 3) the other point of intersection.

Graph:

xy = 12 in green
2x+y = k in blue, where k = 11
1.5 = 3/2

Here's a more zoomed-out look of the graph.

I recommend using either Desmos or GeoGebra as a graphing tool.