Question

For what value of k do the equations $kx-2y=3\mathrm{and}3\mathrm{x}+\mathrm{y}=5$ represent two lines intersecting at a unique point?

(a) k = 3
(b) k = −3
(c) k = 6
(d) all real values except −6

Answer 1

The correct option is (d).

The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁ = −2, c₁ = −3 and a₂ = 3, b₂ = 1 and c₂ = −5
∴ $\frac{a_1}{a_2}=\frac{k}{3},\frac{b_1}{b_2}=\frac{-2}{1}\mathrm{and}\frac{c_1}{c_2}=\frac{-3}{-5}=\frac{3}{5}$
Thus, for these graph lines to intersect at a unique point, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
⇒ $\frac{k}{3}\ne \frac{-2}{1}\Rightarrow k\ne -6$
Hence, the graph lines will intersect at all real values of k except −6.