Correct Answer – Option 3 : 9

Concept:

If a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are 3 lines then these lines are said to be concurrent if:

 \(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

Calculation:

Given:

the lines x + 2y + 1 = 0, 8x + 12y + k = 0 and 3x – 2y + 5 = 0 are concurrent.

Now, by comparing the three lines with the standard equation of line ax + by + c = 0 we get:

⇒ a1 = 1, b1 = 2, c1 = 1, a2 = 8, b2 = 12, c2 = k, a3 = 3, b3 = -2 and c3 = 5.

As we know that, if a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are 3 lines then these lines are said to be concurrent if:

 \(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

\(\Rightarrow \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&2&{1}\\ 8&{12}&k\\ 3&-2&{5} \end{array}} \right| \)

= 1(60 + 2k) – 2(40 – 3k) + (-16 – 36) = 0

⇒ 60 + 2k – 80 + 6k – 52 = 0

8k = 72

k = 9