Straight Lines
Straight Lines
nta_abhyas_2025
Grade 11
Question:
The equation of locus of the foot of perpendicular drawn from $(5,6)$ on the family of lines $(x - 2) + \lambda(y - 3) = 0$ (where $\lambda \in \mathbb{R}$) is
$(x - 1)(x - 3) + (y - 2)(y - 6) = 0$
(x - 5)(x - 6) + (y - 2)(y - 3) = 0$
$(x - 2)(x - 5) + (y - 3)(y - 6) = 0$
$(x + 2)(x + 5) + (y + 3)(y + 6) = 0$
Step-by-Step Solution
Key Concept: To verify if a parametric point lies on a line, substitute its coordinates directly into the line equation and solve for the parameter.
Let the line equation be $\sqrt{3}x - 4y + 8 = 0$.
Let the parametric coordinates be $x = \sqrt{3} - \frac{\sqrt{3}}{2}t$ and $y = 2 - \frac{1}{2}t$.
Substitute these coordinates into the line equation:
$$ \sqrt{3}\left(\sqrt{3} - \frac{\sqrt{3}}{2}t\right) - 4\left(2 - \frac{1}{2}t\right) + 8 = 0 $$
Expand the expression:
$$ 3 - \frac{3}{2}t - 8 + 2t + 8 = 0 $$
Combine the constant terms and the terms involving $t$:
$$ (3 - 8 + 8) + \left(-\frac{3}{2}t + 2t\right) = 0 $$
$$ 3 + \left(\frac{4}{2}t - \frac{3}{2}t\right) = 0 $$
$$ 3 + \frac{1}{2}t = 0 $$
Solve for $t$:
$$ \frac{1}{2}t = -3 $$
$$ t = -6 $$
Correct Answer: -6