Question

What will be the value of k, if the lines given by 3x+ky-4 and 5x+(9+k)y+41 represent two lines intersecting at a point?

(a) k≠\(\frac {7}{2}\)

(b) k≠\(\frac {27}{8}\)

(c) k=\(\frac {27}{2}\)

(d) k≠\(\frac {27}{2}\)

I had been asked this question in an interview for job.

The origin of the question is Solution of Two Linear Equations in Two Variables in Different Methods topic in section Pair of Linear Equations in Two Variables of Mathematics – Class 10

Answer 1

The correct answer is (d) k≠\(\frac {27}{2}\)

To explain I would say: The given equations are 3x+ky-4 and 5x+(9+k)y+41 .

Here, a1=3, b1=k, c1=-4 and a2=5, b2=9+k, c2=41

Lines are intersecting at a point, so \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \)

Now, \(\frac {a_1}{a_2} =\frac {3}{5}, \frac {b_1}{b_2} = \frac {k}{9+k}, \frac {c_1}{c_2} =\frac {-4}{41}\)

\(\frac {3}{5} \ne \frac {k}{9+k}\)

3(9+k)≠5k

27+3k≠5k

27≠5k-3k

2k≠27

k≠\(\frac {27}{2}\)