Question

How many integral values of k are possible if the lines 4x+5ky+7=0 and kx-6y+12=0 intersect in the 2nd quadrant?

Answer 1

Answer:

The total number of positive values of k is 7.

Step-by-step explanation:

Given that

4x+5ky+7=0      ---------(1)

kx-6y+12=0          ------------(2)

If we want to find the value of k then we need to solve the above equation

Now by multiple by k in equation (1)

$4kx+5k^2y+7=0$

kx-6y+12=0

So now by solving above two equation we can find that

$y=\dfrac{48-7k}{6k+5k^2}$

Given that these two lines should cut in the second quadrant ,it means that the value of y should be positive and the value of x should be negative

So we can say that

$\dfrac{48-7k}{6k+5k^2}>0$

⇒   48>7 k

It means that the value of integer k can be lies from negative infinite to 6.It means that k∈[-∞,6].

But The  positive and integer value of  k can be 0,1,2,3,4,5,6.So the total number of positive values of k is 7.