vjsharma25 wrote:

How many integral values of k are possible,if the lines

3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A) 5 B) 4 C) 3 D) 6 E) 2

I was able to do it but I am not convinced by my approach.

Please explain!!

We are given

\(3x+4ky = -6\) .... (1) and

\(kx-3y = -9\) ........(2)

Lets calculate their intersection point in terms of k

Multiply (1) by k and (2) by 3 and then subtract (2) from (1) to solve for y, we will get

\((4k^2+9)*y = 27-6k\) or \(y = (27-6k)/(4k^2+9)\).... (3)

Now, multiply (2) by 4/3k and then add (1) and (2) to solve for x, we will get

\(x = -3*(12k+6)/(4k^2+9)\).... (4)

We know that these intersection points lie in 2nd quadrant, so y has be positive and x has to be negative

y is positive when \(27-6k > 0\)or \(k < 4.5\)

Similarly, x is negative when \(12+6k > 0\) or \(k > -0.5\)

So, we have -0.5<k<4.5, so it can take integer values of 0,1,2,3,4. Hence, five values.

This is tedious approach and takes at least 2.5 to 3 minutes.

I don't know of a faster way to do so