# Two sets, P & Q, satisfy the following conditions P={(x,y):y=2x+5}, Q={(x,y):y=mx+c} & P ∩ Q= Empty set. White down the value of m..& a possible value for c.

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If P ∩ Q= Empty set, the sets don' have any elements in common. That means that the system formed from the equations:

y=2x+5

and

y = mx + c

doesn't have any solution.

The geometric explanation is that the lines whose equations are y=2x+5 and y = mx + c do not have any intercepting point.

We'll re-write the equations:

2x - y = -5 (1)

mx - y = -c (2)

We'll multiply (1) by -m and (2) by 2:

-2mx + my = 5m (3)

2mx - 2y = -2c (4)

We'll add (3) + (4):

-2mx + my + 2mx - 2y = 5m - 2c

We'll eliminate like terms:

my - 2y = 5m - 2c

We'll factorize by y to the left side:

y(m-2) = 5m - 2c

We'll divide by m-2:

y = (5m - 2c)/(m-2)

It is obvious that the ratio doesn't exist if the denominator is cancelling. So, y has no value and also x!

For the denominator m-2 to be zero, we'll form the equation:

m - 2 = 0

We'll add 2:

m = 2

**So, P ∩ Q= Empty set, if and only if m = 2.**