Question

If the system of equations \[ x-2y+z=-4,\qquad 6x+y+kz=15,\qquad 39x-13y+hz=39 \] has infinitely many solutions, then the locus of the point \((h,k)\) is a straight line. The \(x\)-intercept of that line is

• A. 39
• B. 12
• C. 9
• D. 13

Hint:

If a system of three linear equations has infinitely many solutions, then one equation must be a linear combination of the other two. \[ \boxed{ R_3=\lambda R_1+\mu R_2 } \] Comparing corresponding coefficients quickly gives the required relations.

Answer 1

The Correct Option is C

Step 1: Condition for infinitely many solutions. For the given system to have infinitely many solutions, \[ \operatorname{Rank}(A)=\operatorname{Rank}(A|B)<3. \] Since the first two equations are independent, the third equation must be a linear combination of the first two. Let \[ R_3=\lambda R_1+\mu R_2. \]

Step 2: Determine the constants \(\lambda\) and \(\mu\). Comparing the coefficients of \(x\) and \(y\), \[ \lambda+6\mu=39, \] \[ -2\lambda+\mu=-13. \] Solving, \[ \boxed{\lambda=9,\qquad \mu=5.} \]

Step 3: Find the relation between \(h\) and \(k\). Comparing the coefficients of \(z\), \[ h=\lambda+k\mu. \] Substituting the values of \(\lambda\) and \(\mu\), \[ \boxed{h=9+5k.} \] Thus, the locus of \((h,k)\) is \[ \boxed{h-5k-9=0.} \]

Step 4: Find the \(x\)-intercept. The \(x\)-intercept is obtained by putting \[ k=0. \] Hence, \[ h=9. \] Therefore, the \(x\)-intercept is \[ \boxed{9.} \] Hence, \[ \boxed{(C)} \] is the correct answer.