The two pairs of straight lines $12 x^2+7 x y-12 y^2=0$ and $12 x^2+7 x y-12 y^2-x+7 y-1=0$ constitute a
The two pairs of straight lines $12 x^2+7 x y-12 y^2=0$ and $12 x^2+7 x y-12 y^2-x+7 y-1=0$ constitute a
- area of square $\frac{1}{25}$ sq units
- area of square $\frac{1}{5}$ sq units
- area of rectangle $\frac{1}{10}$ sq units
- area of rectangle $\frac{1}{15}$ sq units
Solution
$
\begin{aligned}
& \text { } 12 x^2+7 x y-12 y^2=0 \\
& 12 x^2+16 x y-9 x y-12 y^2=0 \\
& 4 x(3 x+4 y)-3 y(3 x+4 y)=0 \\
& (4 x-3 y)(3 x+4 y)=0 \\
& 4 x-3 y=0 \text { and } 3 x+4 y=0
\end{aligned}
$
Slope of the two lines are $\frac{4}{3}$ and $\frac{-3}{4}$, respectively.
The two lines are perpendicular.
$
\begin{aligned}
& 12 x^2+7 x y-12 y^2-x+7 y-1=0 \\
& 12 x^2+x(7 y-1)-12 y^2+7 y-1=0 \\
& x=-\frac{b+\sqrt{b^2-4 a c}}{2 a} \\
& x=-\frac{(7 y-1) \pm \sqrt{(7 y-1)^2-4 \times 12\left(-12 y^2+7 y-1\right)}}{2 \times 12} \\
& x=\frac{1-7 y \pm \sqrt{49 y^2+1-14 y-336 y+48+576 y^2}}{24} \\
& \Rightarrow 24 x=1-7 y \pm \sqrt{625 y^2-350 y+49} \\
& \Rightarrow 24 x=1-7 y \pm(25 y-7) \\
& \Rightarrow 24 x=1-7 y+25 y-7 \text { or } 24 x=1-7 y-25 y+7 \\
& \Rightarrow 24 x-18 y+6=0 \text { or } 24 x+32 y-8=0 \\
& m_3=\frac{24}{18} \text { and } m_4=\frac{-24}{32} \\
& \Rightarrow m_3=\frac{4}{3} \text { and } m_4=\frac{-3}{4}
\end{aligned}
$
Again, $m_3 \times m_4^*=-1$, the two lines arc perpendicular.
The two lines $3 x+4 y=0$ and $4 x-3 y=0$ intersect at a point $(0,0)$.
$d_1$, and $d_2$ are the distance of a point $(0,0)$ to the line $24 x-18 y+6=0$ and $24 x+32 y-8=0$
$
\begin{aligned}
& d_1=\frac{6}{\sqrt{(24)^2+(18)^2}}, d_2=\frac{8}{\sqrt{(24)^2+(32)^2}} \\
& d_1=\frac{6}{30}=\frac{1}{5}, d_2=\frac{8}{40}=\frac{1}{5}
\end{aligned}
$
$\therefore$ The given figure is a square of side $1 / 5$ unit.
Area of square $=\frac{1}{25}$ sq. units
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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