Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
Let the lines $3x-4y-\alpha = 0$, $8x-11y-33 = 0$, and $2x-3y+\lambda = 0$ be concurrent. If the image of the point $(1,2)$ in the line $2x-3y+\lambda = 0$ is $\left(\dfrac{57}{13},-\dfrac{40}{13}\right)$, then $|\alpha\lambda|$ is equal to
Step-by-Step Solution
Key Concept: Use the reflection midpoint condition to find $\lambda$, then use concurrency (zero determinant) to find $\alpha$, and compute $|\alpha\lambda|$.
Midpoint of $(1,2)$ and $\left(\tfrac{57}{13},-\tfrac{40}{13}\right)$: $M = \left(\tfrac{35}{13},-\tfrac{7}{13}\right)$.
$M$ lies on $2x-3y+\lambda=0$: $\tfrac{70}{13}+\tfrac{21}{13}+\lambda=0 \Rightarrow \lambda=-7$.
Concurrency condition:
$$\begin{vmatrix}3&-4&-\alpha\\8&-11&-33\\2&-3&-7\end{vmatrix}=0.$$
Expanding: $3(-11\cdot(-7)-(-33)(-3))-(-4)(8(-7)-(-33)(2))+(-\alpha)(8(-3)-(-11)(2))=0$.
$\Rightarrow -\lambda+2\alpha-33=0 \Rightarrow 7+2\alpha=33 \Rightarrow \alpha=13$.
$$|\alpha\lambda| = |13\times(-7)| = 91.$$
Correct Answer: 3