First slide

Equation of line in Straight lines

Question

$Consider the family of lines (x - y - 6) + λ (2 x + y + 3) = 0 and (x + 2 y - 4) + μ (3 x - 2 y - 4) = 0$ .If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

Moderate

A

$x^{2} + y^{2} - 3 x + 4 y - 3 = 0$
B

$x^{2} + y^{2} + 3 x - 4 y - 3 = 0$
C

$x^{2} + y^{2} - 3 x - 4 y - 3 = 0$
D

$x^{2} + y^{2} + 3 x + 4 y - 3 = 0$

Solution

$s o l v i n g x - y - 6 = 0 a n d 2 x + y + 3 = 0 w e g e t (1, - 5) and x + 2 y - 4 = 0, 3 x - 2 y - 4 = 0 w e g e t (2, 1)$
Given family of lines are concurrent at A(1, -5) and B(2, 1). Now point P(h, k) is such that AP $⊥$ BP
$\begin{array}{l} \Rightarrow (\frac{k + 5}{h - 1}) \cdot (\frac{k - 1}{h - 2}) = - 1 \\ ∴ Locus of (h, k) is x^{2} + y^{2} - 3 x + 4 y - 3 = 0 . \end{array}$

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