Through the point (2,-3) and parallel to lines represented by x^2… - Vidyadip

Question

Through the point $(2,-3)$ and parallel to lines represented by $x^2+x y-y^2=0$

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Answer

Comparing the equation$x^2+x y-y^2=0 \ldots (1)$
with $a x^2+2 h x y+b y^2=0$, we get,
$a = 1, 2h = 1, b = -1$ Let $m_1$ and $m_2$ be the slopes of the lines represented by $(1)$.
$\left.\begin{array}{l}\text { Then } m_1+m_2=-\frac{2 h}{b}=\frac{-1}{-1}=1 \\ \text { and } m_1 m_2=\frac{a}{b}=\frac{1}{-1}=-1\end{array}\right\} \ldots (2)$
The slopes of the lines parallel to these lines are $m_1$ and $m_2.$
$\therefore $ the equations of the lines with these slopes and through the point $(2, -3$) are
$y + 3 = $m_1$(x – 2)$ and $y + 3 = m_2(x – 2)$
$i.e. m_1(x – 2) – (y + 3) = 0$ and $m_2(x – 2) – (y + 3) = 0$
$\therefore $ the joint equation of these lines is
$[m_1(x – 2) – (y + 3)][m_2(x – 2) – (y + 3)] = 0$
$ \therefore m_1 m_2(x-2)^2-m_1(x-2)(y+3)-m_2(x-2)(y+3)+(y+3)^2=0$
$ \therefore m_1 m_2(x-2)^2-\left(m_1+m_2\right)(x-2)(y+3)+(y+3)^3=0$
$ \therefore(x-2)^2-(x-2)(y+3)+(y+3)^2=0 \ldots \ldots[\text { By (2)] }$
$ \therefore(x-2)^2+(x-2)(y+3)-(y+3)^2=0$
$ \therefore\left(x^2-4 x+4\right)+(x y+3 x-2 y-6)-\left(y^2+6 y+9\right)=0$
$ \therefore x^2-4 x+4+x y+3 x-2 y-6-y^2-6 y-9=0$
$ \therefore x^2+x y-y^2-x-8 y-11=0 .$

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