Q86 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics
The equations of the lines perpendicular to $x^2-5 x y+4 y^2=0$ and passing through $(2,1)$ is
- A$4 x^2+5 x y+y^2-13 x-1=0$
- B$4 x^2+5 x y+y^2-5 x-10 y-7=0$
- C$4 x^2+5 x y+y^2-4 x-4 y-15=0$
- Correct answerD$4 x^2+5 x y+y^2-21 x-12 y+27=0$
Solution
Given, lines are $x-4 y=0$ and $x-y=0$
Lines perpendicular to them and passing through
$(2,1)$ are $4 x+y-9=0$ and $x+y-3=0$
$\therefore$ Equation of pair of straight lines is
$$
\begin{aligned}
& (4 x+y-9)(x+y-3)=0 \\
& \Rightarrow 4 x^2+5 x y+y^2-21 x-12 y+27=0
\end{aligned}
$$
Hence, the correct answer is option: 'D'.