Question

The value of k so that the lines \(x + ky + 2 = 0,\:3x + 2y + 1 = 0\)

\(2x + y + 2 = 0\) from a triangle is
1. \(k \in \left\{ {\frac{2}{5},\frac{2}{3},\frac{3}{4}} \right\}\)
2. \(k \in \left\{ {\frac{1}{4},\frac{2}{3},\frac{3}{4}} \right\}\)
3. \(k \in \left\{ {\frac{1}{4},\frac{2}{3},\frac{1}{2}} \right\}\)
4. \(k \in R-\left\{ {\frac{1}{4},\frac{2}{3},\frac{1}{2}} \right\}\)

Answer 1

Correct Answer - Option 4 : \(k \in R-\left\{ {\frac{1}{4},\frac{2}{3},\frac{1}{2}} \right\}\)

\(- \frac{1}{k} ≠ - \frac{3}{2} \Rightarrow k ≠ \frac{2}{3}\)

\(- \frac{1}{k} - 2 \Rightarrow k ≠ \frac{1}{2}\)

\(\left| {\begin{array}{*{20}{c}} 1&k&2 \\ 3&2&1 \\ 2&1&2 \end{array}} \right| ≠ 0\)

(1) (4 – 1) – k(6 – 2) + 2(–1) ¹ 0

3 – 4k – 2 ≠ 0

k ≠ \(\frac{1}{4}\)

lines will form Δ if is,

k ∈ R – \(\left\{ {\frac{1}{4},\frac{2}{3},\frac{1}{2}} \right\}\)