Question

if lines x+2y+7=0 and 2x+ky+18=0 intersect at a point then find the value of k

Answer 1

a1=1,b1=2,c1=7
a2=2,b2=k,c2=18
Condition for intersecting=
a1/a2≠b1/b2
1/2≠2/k
k≠4
Hence, the value of kis other than 4.

Answer 2

Answer:

The value of k is any real number except 4.

Step-by-step explanation:

Given : If lines x+2y+7=0 and 2x+ky+18=0 intersect at a point.

To find : The value of k?

Solution :

When the system of equation are in form,

$a_1x+b_1y+c_1=0, a_2x+b_2y+c_2=0$

Intersection means they have unique solution satisfying condition,

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

On comparing,

$a_1=1,a_2=2, b_1=2, b_2=k$

Substitute in condition,

$\frac{1}{2}\neq \frac{2}{k}$

Cross multiply,

$k\neq 4$

So, The value of k is any real number except 4.