Matrices & Determinants
System of Equations — Concurrent Lines and Infinite Solutions
nta_pyq_2024_jan
Grade 12
Question:
Let for any three distinct consecutive terms $a,b,c$ of an A.P., the lines $ax+by+c=0$ be concurrent at the point $P$ and $Q(\alpha,\beta)$ be a point such that the system of equations $x+y+z=6$, $2x+5y+\alpha z=\beta$ and $x+2y+3z=4$, has infinitely many solutions. Then $(PQ)^2$ is equal to
Step-by-Step Solution
Key Concept: For AP: $2b=a+c\Rightarrow a-2b+c=0$, so $ax+by+c=0$ passes through fixed point $(1,-2)$, hence $P=(1,-2)$. Then set determinant conditions for infinite solutions of the system to find $Q=(\alpha,\beta)$.
$a,b,c$ in AP: $2b=a+c\Rightarrow a(1)+b(-2)+c=0$, so $P=(1,-2)$. For infinite solutions of the system: $D=\begin{vmatrix}1&1&1\\2&5&\alpha\\1&2&3\end{vmatrix}=0\Rightarrow\alpha=8$. $D_1=0\Rightarrow\beta=6$. $Q=(8,6)$. $(PQ)^2=(7)^2+(8)^2=113$.
Correct Answer: 113