Question of The Day08-06-2020

# If the lines 7x - 3y = 5, 2x – 6y = 6 and 5x + Ky = 4 are concurrent, then what is the value of k?

Correct Answer : a ) 3.875

Explanation :

The lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, and a3x + b3y + c3 = 0 will be concurrent to each other if and only if

$$\begin{bmatrix} a1 & b1 & c1 \\ a2 & b2 & c2 \\ a3 & b3 & c3 \end{bmatrix} =0$$

So, lines 7x - 3y = 5, 2x – 6y = 6 and 5x + Ky = 4 are concurrent to each other if

$$\begin{bmatrix} 7 & -3 & -5 \\ 2 & -6 & -6 \\ 5 & k & -4 \end{bmatrix} =0$$

⇒ 7 (24 – (-6k)) – (- 3) (-8 – (- 30)) + (- 5) (2k – (- 30)) = 0

⇒ 168 + 42k – 24 + 90 – 10k – 150 = 0

⇒ 124 + 32k = 0

⇒ k = – 3.875

Hence, (a) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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