Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 30 Question 30 of 472 Medium MCQ
Let the system of equations :
2 x + 3 y + 5 z = 9 7 x + 3 y − 2 z = 8 12 x + 3 y − ( 4 + λ ) z = 16 − μ \begin{gathered} 2x+3y+5z=9 \\ 7x+3y-2z=8 \\ 12x+3y-(4+\lambda )z=16-\mu \end{gathered} 2 x + 3 y + 5 z = 9 7 x + 3 y − 2 z = 8 12 x + 3 y − ( 4 + λ ) z = 16 − μ
have infinitely many solutions. Then the radius of the circle centred at ( λ , μ ) (\lambda ,\mu ) ( λ , μ ) and touching the line 4 x = 3 y 4x=3y 4 x = 3 y is :
Solutions · 1 Post your solution Official solution
Δ = ∣ 2 3 5 7 3 − 2 12 3 − ( 4 + λ ) ∣ = 2 ( − 12 − 3 λ + 6 ) − 3 ( − 28 − 7 λ + 24 ) + 5 ( 21 − 36 ) = − 12 − 6 λ + 12 + 21 λ − 75 = 15 λ − 75 ⇒ 15 λ − 75 = 0 ⇒ λ = 5 Δ 1 = ∣ 9 3 5 8 3 − 2 16 − μ 3 − 9 ∣ = 9 ( − 27 + 6 ) − 3 ( − 72 + 32 − 2 μ ) + 5 ( 24 − 48 + 3 μ ) = − 189 + 120 + 6 μ − 120 + 15 μ = 21 μ − 189 = 0 ⇒ μ = 9 \begin{gathered} \Delta=|\begin{array}{|c|}2 \quad 3 \quad 5 \\ 7 \quad 3 \quad -2 \\ 12 \quad 3 \quad -(4+\lambda )\end{array}| \\ =2(-12-3\lambda +6)-3(-28-7\lambda +24)+5(21-36) \\ =-12-6\lambda +12+21\lambda -75 \\ =15\lambda -75 \\ \Rightarrow 15\lambda -75=0 \\ \Rightarrow \lambda =5 \\ {\Delta}_{1}=|\begin{array}{|c|}9 \quad 3 \quad 5 \\ 8 \quad 3 \quad -2 \\ 16-\mu \quad 3 \quad -9\end{array}| \\ =9(-27+6)-3(-72+32-2\mu )+5(24-48+3\mu ) \\ =-189+120+6\mu -120+15\mu \\ =21\mu -189=0 \\ \Rightarrow \mu =9 \end{gathered} Δ = ∣ 2 3 5 7 3 − 2 12 3 − ( 4 + λ ) ∣ = 2 ( − 12 − 3 λ + 6 ) − 3 ( − 28 − 7 λ + 24 ) + 5 ( 21 − 36 ) = − 12 − 6 λ + 12 + 21 λ − 75 = 15 λ − 75 ⇒ 15 λ − 75 = 0 ⇒ λ = 5 Δ 1 = ∣ 9 3 5 8 3 − 2 16 − μ 3 − 9 ∣ = 9 ( − 27 + 6 ) − 3 ( − 72 + 32 − 2 μ ) + 5 ( 24 − 48 + 3 μ ) = − 189 + 120 + 6 μ − 120 + 15 μ = 21 μ − 189 = 0 ⇒ μ = 9
∴ r = ∣ 4 ( 5 ) − 3 ( 9 ) ( 4 ) 2 + ( 3 ) 2 ∣ r = 7 5 \begin{gathered} \therefore r=|\frac{4(5)-3(9)}{\sqrt{(4{)}^{2}+(3{)}^{2}}}| \\ r=\frac{7}{5} \end{gathered} ∴ r = ∣ ( 4 ) 2 + ( 3 ) 2 4 ( 5 ) − 3 ( 9 ) ∣ r = 5 7 Answer the question to unlock